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All questions of Ray Optics and Optical Instruments for Class 12 Exam
When light undergoes refraction, its frequency- a)Increases
- b)Remains same
- c)Decreases
- d)Increases exponentially
Correct answer is option 'B'. Can you explain this answer?
When light undergoes refraction, its frequency
a)
Increases
b)
Remains same
c)
Decreases
d)
Increases exponentially
|
Geetika Shah answered |
Light refracts as it passes through a material. It's direction changes, but it still passes through. Light travels at a maximum speed -- the speed of light in a vacuum, but when traveling in anything else it slows down. Different materials slow the speed of passing light at different rates. This property of matter is called the refractive index. When refracting, light doesn't change it's frequency, but since it changes it's speed, it must also change it's wavelength (it gets squished or elongated). Frequency, wavelength, and speed are all related, so if one property changes, another must as well.
When light undergoes refraction, the wavelength- a)Increases in rarer medium
- b)Decreases in rarer medium
- c)Increases in denser medium
- d)Decreases in denser medium
Correct answer is option 'D'. Can you explain this answer?
When light undergoes refraction, the wavelength
a)
Increases in rarer medium
b)
Decreases in rarer medium
c)
Increases in denser medium
d)
Decreases in denser medium
|
Rahul Bansal answered |
When light undergoes refraction at the surface of separation of two media, wavelength decreases on entering a denser medium and wavelength increases on entering a rarer medium.
A convex and a concave mirror of radii 10 cm each are facing each other and 15 cm apart. A point object is placed midway between them. Then position of the final image if the reflection first takes place at the concave mirror and then in the convex mirror is- a)at the pole of the concave mirror
- b)at the pole of the convex mirror
- c)5 cm behind the convex mirror
- d)coincident with the object itself
Correct answer is option 'B'. Can you explain this answer?
A convex and a concave mirror of radii 10 cm each are facing each other and 15 cm apart. A point object is placed midway between them. Then position of the final image if the reflection first takes place at the concave mirror and then in the convex mirror is
a)
at the pole of the concave mirror
b)
at the pole of the convex mirror
c)
5 cm behind the convex mirror
d)
coincident with the object itself
|
Aayush Agarwal answered |
It will be best understood if you make a scale diagram. I will advice you do so and simultaneously read the answer. at first the reflection takes place at concave mirror. the objeCt is at a distance of 7.5cm from it. and 2.5cm from its focus. thereby the image shall be made in front of the mirror. next the image made will act as an object for the convex Mirror. and then it will make the image at its pole. if you arent still able to understand, do let me know
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will be- a)20.2cm
- b)78.23 cm
- c)7.23 cm
- d)2.02 cm
Correct answer is option 'B'. Can you explain this answer?
The focal length of a convex lens (refractive index = 1.5) in air is 20 cm. When immersed in water (refractive index = 1.33), its focal length will be
a)
20.2cm
b)
78.23 cm
c)
7.23 cm
d)
2.02 cm
|
Nikita Singh answered |
Focal length in air = 20 cm
Refractive index of air-water n₁= 1.33
Refractive index of air - glass n₂= 1.5
For focal length in air,
Using formula of lens
1/fair={(n2/n1)-1}(1/R1)-(1/R2)
Put the value into the formula
1/20={(1.5/1)-1}{(1/R1)-(1/R2)}
1/20=0.5{(1/R1)-(1/R2)}…1
We need to calculate the focal length in water
Using formula of lens
1/fwater={(1.5/1.33)-1}{(1/R1)-(1/R2)}
1/fwater=0.128{(1/R1)-(1/R2)}….2
fwater/20=0.5/0.128
fwater=78.125cm
Refractive index of air-water n₁= 1.33
Refractive index of air - glass n₂= 1.5
For focal length in air,
Using formula of lens
1/fair={(n2/n1)-1}(1/R1)-(1/R2)
Put the value into the formula
1/20={(1.5/1)-1}{(1/R1)-(1/R2)}
1/20=0.5{(1/R1)-(1/R2)}…1
We need to calculate the focal length in water
Using formula of lens
1/fwater={(1.5/1.33)-1}{(1/R1)-(1/R2)}
1/fwater=0.128{(1/R1)-(1/R2)}….2
fwater/20=0.5/0.128
fwater=78.125cm
Due to _______ the depth of an optically denser medium appears to be _______ than its real depth.- a)Refraction , more
- b)Reflection , less
- c)Reflection , more
- d)Refraction , less
Correct answer is option 'D'. Can you explain this answer?
Due to _______ the depth of an optically denser medium appears to be _______ than its real depth.
a)
Refraction , more
b)
Reflection , less
c)
Reflection , more
d)
Refraction , less
|
|
Neha Sharma answered |
When light moves from 1 medium to another, refraction takes place. And when light enters a denser medium, the ray bends away from normal and meets at a point above the actual point where they would have met, if the medium was absent. So, apparent depth decreases.
A thick plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face. Then, x is equal to:
- a)2.5cm
- b)2.3cm
- c)2cm
- d)2.1cm
Correct answer is option 'A'. Can you explain this answer?
A thick plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face. Then, x is equal to:
a)
2.5cm
b)
2.3cm
c)
2cm
d)
2.1cm
|
EduRev Humanities answered |
A thick Plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face, and we have to determine the value of ' x'
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm
Power of the lens is -40, its focal length is - a)4m
- b)-40m
- c)-0.25m
- d)-25m
Correct answer is option 'C'. Can you explain this answer?
Power of the lens is -40, its focal length is
a)
4m
b)
-40m
c)
-0.25m
d)
-25m
|
Jyoti Sengupta answered |
The power of a lens is the reciprocal of the focal length with measurement in metres. The unit is diopter.
It is given that the focal length of a convex lens is 10 cm = 0.1 m.
⇒ The power of the lens is 1/0.1= 10 diopter.
Just before setting, the sun may appear to be elliptical. This happens due to:
- a)Refraction
- b)Dispersion
- c)Reflection
- d)Diffraction
Correct answer is option 'A'. Can you explain this answer?
Just before setting, the sun may appear to be elliptical. This happens due to:
a)
Refraction
b)
Dispersion
c)
Reflection
d)
Diffraction
|
Om Desai answered |
Refraction of light ray through the atmosphere may cause different magnification in mutually perpendicular directions.
Sunlight enters the Earth's atmosphere from the vacuum of space. The refractive index of air with respect to vacuum is 1.0029. This means that the speed of light in air is 1.0029 times slower than the speed of light in vacuum. This causes the light rays to bend towards the normal (refraction).
Sunlight enters the Earth's atmosphere from the vacuum of space. The refractive index of air with respect to vacuum is 1.0029. This means that the speed of light in air is 1.0029 times slower than the speed of light in vacuum. This causes the light rays to bend towards the normal (refraction).
Bending effect of light as it passes from one transparent material into other is known as- a)refraction
- b)deflection
- c)reflection
- d)diffraction
Correct answer is option 'A'. Can you explain this answer?
Bending effect of light as it passes from one transparent material into other is known as
a)
refraction
b)
deflection
c)
reflection
d)
diffraction
|
|
Leelu Bhai answered |
Its refraction yrr.. This is a common definition of refraction that u r asking 🤦🏻♂️
An air bubble inside a glass slab (μ = 1.5) appears at 6 cm when viewed from the opposite side. The thickness of the slab is:- a)10 cm
- b)6.67 cm
- c)15 cm
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
An air bubble inside a glass slab (μ = 1.5) appears at 6 cm when viewed from the opposite side. The thickness of the slab is:
a)
10 cm
b)
6.67 cm
c)
15 cm
d)
None of the above
|
Krishna Iyer answered |
We know that μ=(apparent depth/real depth)
Let the thickness of the slab be t and real depth of the bubble from one side be x. Then
μ=(x/6)=(t−x)/4 or 1.5=x/6 = (t−x)/4
This gives x=9 and 1.5=(t−9)/4 or t=15cm
A microscope is focused on an ink mark on the top of a table. If we place a glass slab 3cm thick on it, how should the microscope be moved to focus the ink spot again? The refractive index of glass slab is 1.5 cm.- a)2 cm downward
- b)1 cm upwards
- c)1 cm downwards
- d)2 cm upwards
Correct answer is option 'B'. Can you explain this answer?
A microscope is focused on an ink mark on the top of a table. If we place a glass slab 3cm thick on it, how should the microscope be moved to focus the ink spot again? The refractive index of glass slab is 1.5 cm.
a)
2 cm downward
b)
1 cm upwards
c)
1 cm downwards
d)
2 cm upwards
|
Shreya Gupta answered |
The image of the ink spot moves up by 1cm.
Refractive index = Real depth/ apparent depth
1.5 = 3/ apparent depth
Apparent depth = 3/1.5 = 2 cm
Distance through which the microscope is moved is 3 cm -2 cm = 1 cm upwards
______ mirror is called as diverging mirror- a)Concave
- b)Plane
- c)Convex
- d)Both b and c
Correct answer is option 'C'. Can you explain this answer?
______ mirror is called as diverging mirror
a)
Concave
b)
Plane
c)
Convex
d)
Both b and c
|
Preeti Iyer answered |
Concave mirror is called a converging mirror because parallel rays of light fall on the mirror they converge at a point called focus. Convex mirror is called a diverging mirror because parallel rays of light fall on it they diverge after reflection.
What happens to the magnifying power of microscope, when its length increases?- a)May increase or decrease depending on the relative focal lengths of objective and eye piece.
- b)decreases
- c)Unchanged
- d)Increases
Correct answer is option 'D'. Can you explain this answer?
What happens to the magnifying power of microscope, when its length increases?
a)
May increase or decrease depending on the relative focal lengths of objective and eye piece.
b)
decreases
c)
Unchanged
d)
Increases
|
Hansa Sharma answered |
Magnification in microscope is m=− lD/f0fe
where, l is length of tube and D is least distance for clear vision
f0 is focal length of objective
fe is the focal length of the eyepiece.
So, as l increases m increases.
Hence, the magnifying power of a microscope increases with increase in tube length.
where, l is length of tube and D is least distance for clear vision
f0 is focal length of objective
fe is the focal length of the eyepiece.
So, as l increases m increases.
Hence, the magnifying power of a microscope increases with increase in tube length.
A prism of refractive index √2 and refractive angle A produces minimum deviation δm of a ray on one face at an angle of incidence 45°. The values of A and δm are respectively- a)45°, 45°
- b)60°, 30°
- c)60°, 45°
- d)45°, 60°
Correct answer is option 'B'. Can you explain this answer?
A prism of refractive index √2 and refractive angle A produces minimum deviation δm of a ray on one face at an angle of incidence 45°. The values of A and δm are respectively
a)
45°, 45°
b)
60°, 30°
c)
60°, 45°
d)
45°, 60°
|
Bs Academy answered |
μ=sinisinr
Again for minimum deviation
μ=sin(A+δmin2)sinA2
∴μ=sinisinA2
⇒sinA2=siniμ
⇒sinA2=sin45∘√2=12
⇒A2=30∘
⇒A=60∘
Again i+e=A+δmin
For minimum deviation i=e
∴2i=A+δmin
⇒δmin=2i−A
⇒δmin=2×45∘−60∘=30∘
A ray of light going from denser to rarer medium suffers refraction at a concave surface. Which of the following relations is correct?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A ray of light going from denser to rarer medium suffers refraction at a concave surface. Which of the following relations is correct?
a)
b)
c)
d)
|
EduRev JEE answered |
Solution :
The correct option is Option A.
Laws of refraction;-
The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.
For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.
When a ray of light passes through a glass slab, ∠i,∠r and the normal all lie in the same plane.
When a ray of light passes from one medium to another, here from air to glass or glass to air, the ratio sini / sinr = constant.
A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation . The angle of prism P2 is:- a)5.33°
- b)2.6°
- c)3°
- d)4°
Correct answer is option 'C'. Can you explain this answer?
A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation . The angle of prism P2 is:
a)
5.33°
b)
2.6°
c)
3°
d)
4°
|
|
Rajat Kapoor answered |
If R1 and R2 are the radii of curvature of a double convex lens, which of the following will have the largest power?- a)R1 = 10 cm and R2 = ∞
- b)R1 = R2 = 5 cm
- c)R1 = ∞ and R2 = 10 cm
- d)R1 = R2 = 10 cm
Correct answer is option 'B'. Can you explain this answer?
If R1 and R2 are the radii of curvature of a double convex lens, which of the following will have the largest power?
a)
R1 = 10 cm and R2 = ∞
b)
R1 = R2 = 5 cm
c)
R1 = ∞ and R2 = 10 cm
d)
R1 = R2 = 10 cm
|
|
Nikita Singh answered |
Using the lens maker’s formula we get
p= (μ−1) (1/R1+1/R2)
As we can clearly see, when R1= R2= 5cm then the maximum power is achieved.
p= (μ−1) (1/R1+1/R2)
As we can clearly see, when R1= R2= 5cm then the maximum power is achieved.
Can absolute value of refractive index be less than unity?- a)No
- b)Refractive index is always unity
- c)Depends on situation
- d)Yes
Correct answer is option 'A'. Can you explain this answer?
Can absolute value of refractive index be less than unity?
a)
No
b)
Refractive index is always unity
c)
Depends on situation
d)
Yes
|
Vijay Bansal answered |
Absolute refractive index of a medium = speed of light in vacuum / speed of light in in that medium.
Since the speed of light is maximum in vacuum, the refractive index cannot be less than unity.
An object is placed at a distance of 10 cm from a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30 cm, what is the nature and focal length of lens A ?- a)Concave, 12 cm
- b)Convex, 18 cm
- c)Convex, 12 cm
- d)Convex, 6 cm
Correct answer is option 'D'. Can you explain this answer?
An object is placed at a distance of 10 cm from a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30 cm, what is the nature and focal length of lens A ?
a)
Concave, 12 cm
b)
Convex, 18 cm
c)
Convex, 12 cm
d)
Convex, 6 cm
|
|
Riya Banerjee answered |
The critical angle for the material of a prism is 45° and its refracting angle is 30°. A monochromatic ray goes out perpendicular to the surface of emergence from the prism. What is the angle of incidence on the prism?- a)60°
- b)75°
- c)30°
- d)45°
Correct answer is option 'D'. Can you explain this answer?
The critical angle for the material of a prism is 45° and its refracting angle is 30°. A monochromatic ray goes out perpendicular to the surface of emergence from the prism. What is the angle of incidence on the prism?
a)
60°
b)
75°
c)
30°
d)
45°
|
Sanchita Iyer answered |
Magnifying power of a compound microscope is high if- a)the objective has a long focal length and eye-piece has a short focal
- b)the objective has a short focal length and the eye-piece has a long
- c)both objective and eye-piece have short focal lengths
- d)both objective and eye-piece have long focal lengths
Correct answer is option 'C'. Can you explain this answer?
Magnifying power of a compound microscope is high if
a)
the objective has a long focal length and eye-piece has a short focal
b)
the objective has a short focal length and the eye-piece has a long
c)
both objective and eye-piece have short focal lengths
d)
both objective and eye-piece have long focal lengths
|
Ciel Knowledge answered |
Magnifying power of a compound- microscope is given by m=− (lD/f0fe)
where, l is length of tube
D is least distance of clear vision
f0 is focal length of objective
fe is focal length of eyepiece
So, clearly, it can be seen that focal length of objective and eyepiece needs to be decreased so that magnifying power increases.
where, l is length of tube
D is least distance of clear vision
f0 is focal length of objective
fe is focal length of eyepiece
So, clearly, it can be seen that focal length of objective and eyepiece needs to be decreased so that magnifying power increases.
A virtual image, larger than the object can be produced by- a)Convex mirror
- b)Concave mirror
- c)Plane mirror
- d)Concave lens
Correct answer is option 'B'. Can you explain this answer?
A virtual image, larger than the object can be produced by
a)
Convex mirror
b)
Concave mirror
c)
Plane mirror
d)
Concave lens
|
|
Ayaan Madhukar answered |
CONCAVE MIRROR...
Concave mirrors produce real and virtual, erect and inverted, diminished same size and magnified image depending upon the position of the object on the principal axis.
But the concave mirror forms a virtual and enlarged image when the object is placed between the focus and the pole of the mirror...
$$Hope it's help...$$
Concave mirrors produce real and virtual, erect and inverted, diminished same size and magnified image depending upon the position of the object on the principal axis.
But the concave mirror forms a virtual and enlarged image when the object is placed between the focus and the pole of the mirror...
$$Hope it's help...$$
Total internal reflection occurs when- a)Angle of incidence is equal to critical angle
- b)Angle of incidence is greater than critical angle
- c)Total internal reflection doesn’t depend on angle of incidence or critical angle
- d)Angle of incidence is less than critical angle
Correct answer is option 'B'. Can you explain this answer?
Total internal reflection occurs when
a)
Angle of incidence is equal to critical angle
b)
Angle of incidence is greater than critical angle
c)
Total internal reflection doesn’t depend on angle of incidence or critical angle
d)
Angle of incidence is less than critical angle
|
Rounak Goyal answered |
Total internal reflection is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface.The critical angle is the angle of incidence above which the total internal reflection occurs.
Light is confined within the core of a simple optical fiber by:- a)total internal reflection at the outer edge of the cladding.
- b)reflection from the fiber’s plastic coating.
- c)total internal reflection at the core cladding boundary.
- d)refraction
Correct answer is option 'C'. Can you explain this answer?
Light is confined within the core of a simple optical fiber by:
a)
total internal reflection at the outer edge of the cladding.
b)
reflection from the fiber’s plastic coating.
c)
total internal reflection at the core cladding boundary.
d)
refraction
|
|
Riya Banerjee answered |
Light remains confined within the core of simple optical fibre because of Total internal reflection from core cladding boundary.
Light is confined within the core of a simple optical fiber by. If light hits a boundary of a material of lower refractive index at a steep enough angle, it cannot get out and it's reflected back into the high index medium, as in the figure below.
An optical fibre is a thin rod of high-quality glass. Very little light is absorbed by the glass.Optical fibres can carry more information than an ordinary cable of the same thickness. The signals in optical fibres do not weaken as much over long distances as the signals in ordinary cables.
Total internal reflection. When light traveling in an optically dense medium hits a boundary at a steep angle (larger than the critical angle for the boundary), the light is completely reflected. This effect is used in optical fibers to confine light in the core.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
Light is confined within the core of a simple optical fiber by. If light hits a boundary of a material of lower refractive index at a steep enough angle, it cannot get out and it's reflected back into the high index medium, as in the figure below.
An optical fibre is a thin rod of high-quality glass. Very little light is absorbed by the glass.Optical fibres can carry more information than an ordinary cable of the same thickness. The signals in optical fibres do not weaken as much over long distances as the signals in ordinary cables.
Total internal reflection. When light traveling in an optically dense medium hits a boundary at a steep angle (larger than the critical angle for the boundary), the light is completely reflected. This effect is used in optical fibers to confine light in the core.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
In a concave mirror when the object is located beyond C the magnification is- a)More than 1
- b)Equal to 1
- c)Less than 1
- d)Both a and b
Correct answer is option 'C'. Can you explain this answer?
In a concave mirror when the object is located beyond C the magnification is
a)
More than 1
b)
Equal to 1
c)
Less than 1
d)
Both a and b
|
|
Rajeev Saxena answered |
In the animation above, a right-side-up object is located above the principal axis at a position beyond the center of curvature (C). The ray diagram shows that the image of this object is located as an upside-down image positioned between the center of curvature (C) and the focal point (F). In fact, it can be generalized that anytime the object is located beyond the center of curvature, the image will be located somewhere between the center of curvature and the focal point. In such cases, the image will be inverted and reduced in size (i.e., smaller than the object). Such images are called real images because they are formed by the actual convergence of reflected light rays at the image location. Real images are always formed on the same side of the mirror as the object.
A ray of light passes through a plane glass slab of thickness t and refractive index μ = 1.5. The angle between the incident ray and emergent ray will be:- a)30°
- b)45°
- c)60°
- d)0°
Correct answer is option 'D'. Can you explain this answer?
A ray of light passes through a plane glass slab of thickness t and refractive index μ = 1.5. The angle between the incident ray and emergent ray will be:
a)
30°
b)
45°
c)
60°
d)
0°
|
Ayush Joshi answered |
The incident ray and emergent ray are parallel to each other but latteray displaced due to reflaction at two surfaces . So, the angle between them is Zero.
When light is reflected from a mirror a change occurs in its- a)phase,
- b)frequency,
- c)wavelength,
- d)speed
Correct answer is option 'A'. Can you explain this answer?
When light is reflected from a mirror a change occurs in its
a)
phase,
b)
frequency,
c)
wavelength,
d)
speed
|
Ramesh Chand answered |
When a light wave is reflected from an object , it changes not only it's amplitude but also its phase according to the properties of the object at a particular point. Therefore, option A is the right answer.
An image is upright and reduced in size. Which mirror is used to form such an image?- a)Convex
- b)Concave
- c)Plane
- d)Both a and c
Correct answer is option 'D'. Can you explain this answer?
An image is upright and reduced in size. Which mirror is used to form such an image?
a)
Convex
b)
Concave
c)
Plane
d)
Both a and c
|
|
Nikita Singh answered |
Only Convex Mirror can form an image which is upright and reduced in size.
A convex lens produces a real image m times the size of the object. What is the distance of the object from the lens?- a)(m + 1)f
- b)
- c)(m – 1)f
- d)
Correct answer is option 'D'. Can you explain this answer?
A convex lens produces a real image m times the size of the object. What is the distance of the object from the lens?
a)
(m + 1)f
b)
c)
(m – 1)f
d)
|
|
Rajesh Gupta answered |
M=−ϑ /u =−m/1
ϑ=−mu
(1/ϑ)−(1/u)=1/f
1/mu+1/u=−1/f
(1+m)/xu=−1/f
u= (1+m)f/(m)
ϑ=−mu
(1/ϑ)−(1/u)=1/f
1/mu+1/u=−1/f
(1+m)/xu=−1/f
u= (1+m)f/(m)
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will- a)become zero
- b)become infinite
- c)become small, but non-zero
- d)remain unchanged
Correct answer is option 'B'. Can you explain this answer?
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will
a)
become zero
b)
become infinite
c)
become small, but non-zero
d)
remain unchanged
|
Lia Ross answered |
Sure. The answer is b). The focal length of a lens is inversely proportional to the refractive index of the medium in which it is placed. When the refractive index of the lens is equal to the refractive index of the liquid, the lens will act like a plane parallel slab, which has an infinite focal length.
Here's an explanation:
The focal length of a lens is determined by the following formula:
f = (n - 1)r / (n + 1)
where:
f is the focal length of the lens
n is the refractive index of the lens material
r is the radius of curvature of the lens
When the refractive index of the lens is equal to the refractive index of the liquid, the term (n - 1) becomes 0. This means that the focal length of the lens becomes infinite.
In other words, when a convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens, it will no longer be able to focus light rays. Instead, the light rays will pass through the lens without being refracted.
This is because the lens is no longer able to bend the light rays, as the refractive index of the lens and the liquid are the same. As a result, the focal length of the lens becomes infinite.
A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be- a)10 cm
- b)15 cm
- c)20 cm
- d)30 cm
Correct answer is option 'B'. Can you explain this answer?
A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be
a)
10 cm
b)
15 cm
c)
20 cm
d)
30 cm
|
|
Hansa Sharma answered |
Given, u=−20cm
m=−3⇒−v/u =−3
v=−60cm
1/f=(1/v)+(1/u)=(-1/60)−(1/20)=−1−3/60
f=−60/4=−15cm
m=−3⇒−v/u =−3
v=−60cm
1/f=(1/v)+(1/u)=(-1/60)−(1/20)=−1−3/60
f=−60/4=−15cm
For a rectangular slab, refraction takes place at- a)Two interfaces
- b)Three interfaces
- c)Four interfaces
- d)One interfaces
Correct answer is option 'A'. Can you explain this answer?
For a rectangular slab, refraction takes place at
a)
Two interfaces
b)
Three interfaces
c)
Four interfaces
d)
One interfaces
|
|
Saanvi Bose answered |
Refraction in a Rectangular Slab
In a rectangular slab, refraction takes place at two interfaces.
Explanation:
- A rectangular slab can be considered as a prism with a rectangular cross-section.
- When a ray of light enters the slab from air (or any other medium), it undergoes refraction at the first interface between air and the slab.
- The angle of incidence and the angle of refraction are related by Snell's law: n1 sinθ1 = n2 sinθ2, where n1 and n2 are the refractive indices of air and the slab, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively.
- The refracted ray travels through the slab and undergoes refraction again when it exits the slab and enters air (or any other medium).
- The angle of incidence and the angle of refraction at the second interface are related by Snell's law in a similar manner.
- Therefore, refraction takes place at two interfaces in a rectangular slab.
Conclusion:
Refraction in a rectangular slab takes place at two interfaces, one when the ray enters the slab and another when it exits the slab.
In a rectangular slab, refraction takes place at two interfaces.
Explanation:
- A rectangular slab can be considered as a prism with a rectangular cross-section.
- When a ray of light enters the slab from air (or any other medium), it undergoes refraction at the first interface between air and the slab.
- The angle of incidence and the angle of refraction are related by Snell's law: n1 sinθ1 = n2 sinθ2, where n1 and n2 are the refractive indices of air and the slab, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively.
- The refracted ray travels through the slab and undergoes refraction again when it exits the slab and enters air (or any other medium).
- The angle of incidence and the angle of refraction at the second interface are related by Snell's law in a similar manner.
- Therefore, refraction takes place at two interfaces in a rectangular slab.
Conclusion:
Refraction in a rectangular slab takes place at two interfaces, one when the ray enters the slab and another when it exits the slab.
A concave mirror cannot form- a)virtual image of virtual object
- b)virtual image of a real object
- c)real image of a real object
- d)real image of a virutal object
Correct answer is option 'A'. Can you explain this answer?
A concave mirror cannot form
a)
virtual image of virtual object
b)
virtual image of a real object
c)
real image of a real object
d)
real image of a virutal object
|
Ritu Singh answered |
-Virtual image cannot be formed from a virtual object.
- When an object is placed between pole and focus, the image formed by the concave mirror is magnified, virtual and erect.
-When an object is placed beyond the centre of curvature , the image of it is formed between the centre of curvature and focus which is diminished,real and inverted,so the real image of the real object can be formed by a concave mirror.
-When the object is virtual, the image formed is real for the concave surface as shown in figure.
- When an object is placed between pole and focus, the image formed by the concave mirror is magnified, virtual and erect.
-When an object is placed beyond the centre of curvature , the image of it is formed between the centre of curvature and focus which is diminished,real and inverted,so the real image of the real object can be formed by a concave mirror.
-When the object is virtual, the image formed is real for the concave surface as shown in figure.
What is dispersive power?- a)The ratio of angular dispersion to the angle of deviation for the mean wavelength.
- b)The ratio of angular dispersion to the angle of refraction for the mean wavelength.
- c)The ratio of angular dispersion to the angle of reflection for the mean wavelength.
- d)The ratio of angular dispersion to the angle of deviation for the total wavelength.
Correct answer is option 'A'. Can you explain this answer?
What is dispersive power?
a)
The ratio of angular dispersion to the angle of deviation for the mean wavelength.
b)
The ratio of angular dispersion to the angle of refraction for the mean wavelength.
c)
The ratio of angular dispersion to the angle of reflection for the mean wavelength.
d)
The ratio of angular dispersion to the angle of deviation for the total wavelength.
|
|
Naina Bansal answered |
Dispersive power of a prism is defined as the ratio between angular dispersion to mean deviation produced by the prism.
If dμ denotes the difference between the refractive indices of material of prism for violet and red light,
ω = δμ / μ – 1
Here ‘μ’ is the refractive index of prism for a mean colour. A mean colour is that colour whose wavelength lies in between that of violet and red. For white light, yellow colour is, generally, taken to be the mean colour.
Since μv is always greater than μr, the dispersive power of a prism is always positive. It depends upon the type of glass used. It is different for crown glass and for flint glass.
The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length will be - a)40 cm
- b)20 cm
- c)80 cm
- d)30 cm
Correct answer is option 'A'. Can you explain this answer?
The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length will be
a)
40 cm
b)
20 cm
c)
80 cm
d)
30 cm
|
|
Hansa Sharma answered |
Let the radius of curvature of the convex lens are denoted as
For a convex lens
R1=+40cm [As it is towards right]
R2=-40cm [As it is towards left]
The refractive index of the lens [µ]=1.5
Let f is the focal length of the lens.
From lens maker's formula we know that –
1/f=[µ-1] [1/R1-1/R2]
1/f=[1.5-1][1/40-(1/-40)]
1/f=0.5 x (2/40)
1/f=1/40
f=40cm
Hence, the focal length of the lens is 40 cm.
For a convex lens
R1=+40cm [As it is towards right]
R2=-40cm [As it is towards left]
The refractive index of the lens [µ]=1.5
Let f is the focal length of the lens.
From lens maker's formula we know that –
1/f=[µ-1] [1/R1-1/R2]
1/f=[1.5-1][1/40-(1/-40)]
1/f=0.5 x (2/40)
1/f=1/40
f=40cm
Hence, the focal length of the lens is 40 cm.
Which is the position of minimum deviation in prism?- a)When the angle of incidence is such that the refracted ray inside the prism is perpendicular to the base of the prism.
- b)When the angle of incidence is such that the refracted ray inside the prism is parallel to the base of the prism.
- c)When the angle of incidence is such that the refracted ray inside the prism reflects back from the base of the prism.
- d)When the angle of incidence is such that the refracted ray inside the prism is at 45 degrees to the base of the prism.
Correct answer is option 'B'. Can you explain this answer?
Which is the position of minimum deviation in prism?
a)
When the angle of incidence is such that the refracted ray inside the prism is perpendicular to the base of the prism.
b)
When the angle of incidence is such that the refracted ray inside the prism is parallel to the base of the prism.
c)
When the angle of incidence is such that the refracted ray inside the prism reflects back from the base of the prism.
d)
When the angle of incidence is such that the refracted ray inside the prism is at 45 degrees to the base of the prism.
|
|
Priyanshu Pandey answered |
A prism is said to be in the position of minimum deviation when the angle of incidence of the light ray at the first prism surface is equal to the angle of emergence at the second surface of prism. In this condition, the refracted ray inside the prism is parallel to the base of the prism.
A convex lens forms a real image of an object on a screen; the magnification of the image being 3/2. The object and the screen are kept fixed and the lens is moved through a distance of 16 cm when a sharp image is again formed on the screen; the magnification now being 2/3. What is the focal length of the lens?- a)19.6 cm
- b)19.2 cm
- c)20 cm
- d)20.4 cm
Correct answer is option 'B'. Can you explain this answer?
A convex lens forms a real image of an object on a screen; the magnification of the image being 3/2. The object and the screen are kept fixed and the lens is moved through a distance of 16 cm when a sharp image is again formed on the screen; the magnification now being 2/3. What is the focal length of the lens?
a)
19.6 cm
b)
19.2 cm
c)
20 cm
d)
20.4 cm
|
|
Gaurav Kumar answered |
3/2=v/u…...1
2/3=(v-16)/(u+16)......2
from these equation
we get v=48, u=32
so now f=(vxu)/(v+u)=96/5=19.2
The correct answer is option B.
2/3=(v-16)/(u+16)......2
from these equation
we get v=48, u=32
so now f=(vxu)/(v+u)=96/5=19.2
The correct answer is option B.
A myopic person can see things clearly if they lie between 8 cm and 100 cm from his eye. The lens will enable him to see the moon distinctly if its focal length is:- a)infinity
- b)-100 cm
- c)zero
- d)+ 100 cm
Correct answer is option 'B'. Can you explain this answer?
A myopic person can see things clearly if they lie between 8 cm and 100 cm from his eye. The lens will enable him to see the moon distinctly if its focal length is:
a)
infinity
b)
-100 cm
c)
zero
d)
+ 100 cm
|
|
Naina Bansal answered |
The final image of the moon should be formed at a distance of 100 cm. Using a concave lens of focal length 100 cm will enable him to see the moon distinctly.
Focal length of lens =-100 cm
The bluish color predominates in a clear sky,- a)Since blue has a is radiated much more strongly
- b)Since blue has a longer wavelength than red and is scattered much more strongly
- c)Since blue has a shorter wavelength than red and is scattered much more strongly
- d)Since blue has a shorter wavelength than red and is scattered much less strongly
Correct answer is option 'C'. Can you explain this answer?
The bluish color predominates in a clear sky,
a)
Since blue has a is radiated much more strongly
b)
Since blue has a longer wavelength than red and is scattered much more strongly
c)
Since blue has a shorter wavelength than red and is scattered much more strongly
d)
Since blue has a shorter wavelength than red and is scattered much less strongly
|
|
Suresh Kumar answered |
Sky is blue because of scattering of light of minimum wavelength i.e blue by the dust particles suspended in the air. so option c is correct
The magnifying power of the telescope can be increased by- a)increasing the focal length of eyepiece
- b)fitting eye-piece of low power
- c)increasing the distance of an object
- d)fitting eye-piece of high power
Correct answer is option 'D'. Can you explain this answer?
The magnifying power of the telescope can be increased by
a)
increasing the focal length of eyepiece
b)
fitting eye-piece of low power
c)
increasing the distance of an object
d)
fitting eye-piece of high power
|
|
Riya Banerjee answered |
Magnifying power of a telescope is f0/fe , so as 1/fe increases, magnifying power increases.
Hence the correct answer is option D.
Hence the correct answer is option D.
A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is- a) red
- b)blue
- c)green
- d)violet
Correct answer is option 'A'. Can you explain this answer?
A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
a)
red
b)
blue
c)
green
d)
violet
|
Gayatri Sharma answered |
Explanation:
When a short pulse of white light is incident from air to a glass slab at normal incidence, it enters the glass slab and undergoes refraction. The different colors of light have different wavelengths and hence, they bend at different angles. This phenomenon is called dispersion of light.
The refractive index of the glass is different for different colors of light. The refractive index for red light is the least and for violet light is the highest. Thus, the red light bends the least and violet light bends the most.
Emerging color:
As the white light enters the glass slab, it gets dispersed into its constituent colors. The red light has the least deviation and hence emerges first. The other colors follow in the order of their deviation, with violet light emerging last.
Thus, the first color to emerge from the glass slab is red.
Conclusion:
Hence, the correct option is (a) red.
When a short pulse of white light is incident from air to a glass slab at normal incidence, it enters the glass slab and undergoes refraction. The different colors of light have different wavelengths and hence, they bend at different angles. This phenomenon is called dispersion of light.
The refractive index of the glass is different for different colors of light. The refractive index for red light is the least and for violet light is the highest. Thus, the red light bends the least and violet light bends the most.
Emerging color:
As the white light enters the glass slab, it gets dispersed into its constituent colors. The red light has the least deviation and hence emerges first. The other colors follow in the order of their deviation, with violet light emerging last.
Thus, the first color to emerge from the glass slab is red.
Conclusion:
Hence, the correct option is (a) red.
Waves on top of spectrum are- a)infrared waves
- b)gamma rays
- c)x-rays
- d)ultraviolet rays
Correct answer is option 'A'. Can you explain this answer?
Waves on top of spectrum are
a)
infrared waves
b)
gamma rays
c)
x-rays
d)
ultraviolet rays
|
|
Naina Bansal answered |
The correct answer is b.
Infrared: Night vision goggles pick up the infrared light emitted by our skin and objects with heat. In space, infrared light helps us map the dust between stars.
Infrared: Night vision goggles pick up the infrared light emitted by our skin and objects with heat. In space, infrared light helps us map the dust between stars.
Visible: Our eyes detect visible light. Fireflies, light bulbs, and stars all emit visible light.
Ultraviolet: Ultraviolet radiation is emitted by the Sun and are the reason skin tans and burns. "Hot" objects in space emit UV radiation as well.
X-ray: A dentist uses X-rays to image your teeth, and airport security uses them to see through your bag. Hot gases in the Universe also emit X-rays.
Gamma ray: Doctors use gamma-ray imaging to see inside your body. The biggest gamma-ray generator of all is the Universe.
The objective of a telescope has a focal length of 1.2 m. It is used to view a 10.0 m tall tower 2 km away. What is the height of the image of the tower formed by the objective?- a)8 mm
- b)2 mm
- c)6 mm
- d)4 mm
Correct answer is option 'C'. Can you explain this answer?
The objective of a telescope has a focal length of 1.2 m. It is used to view a 10.0 m tall tower 2 km away. What is the height of the image of the tower formed by the objective?
a)
8 mm
b)
2 mm
c)
6 mm
d)
4 mm
|
Lavanya Menon answered |
α (Radian)=10/2000=1/200rad
α=h/F=h/1.2
h= α x 1.2
h=(1/200) x1.2 cm
=6mm
α=h/F=h/1.2
h= α x 1.2
h=(1/200) x1.2 cm
=6mm
The ratio of angular dispersion of to the angle of deviation for the mean wavelength is called Dispersive Power. Represented by
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The ratio of angular dispersion of to the angle of deviation for the mean wavelength is called Dispersive Power. Represented by
a)
b)
c)
d)
|
Divey Sethi answered |
The ratio of angular dispersion to the angle of deviation for the mean wavelength is called Dispersive Power. Represented by 
where δv − δr is the angular dispersion for violet and red lights and δr the deviation suffered by the mean light.
Mark the correct options- a)If the incident rays are converging, we have a real object.
- b)If the final rays are converging, we have a real image.
- c)The image of a virtual object is called a virtual image.
- d)If the image is virtual, the corresponding object is called a virtual object
Correct answer is option 'B'. Can you explain this answer?
Mark the correct options
a)
If the incident rays are converging, we have a real object.
b)
If the final rays are converging, we have a real image.
c)
The image of a virtual object is called a virtual image.
d)
If the image is virtual, the corresponding object is called a virtual object
|
|
Geetika Shah answered |
If the final rays are converging, we have a real image.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
The correct answer is option B.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
The correct answer is option B.
We have a right angled isosceles prism, Its refractive index is 1.5. If we incident a ray normally on one of the two perpendicular surfaces, which of the following phenomenon will take place?- a)Dispersion
- b)Total internal reflection
- c)Refraction
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
We have a right angled isosceles prism, Its refractive index is 1.5. If we incident a ray normally on one of the two perpendicular surfaces, which of the following phenomenon will take place?
a)
Dispersion
b)
Total internal reflection
c)
Refraction
d)
None of the above
|
|
Hansa Sharma answered |
Critical angle will be, sin−1(1/1.5) = 41.8. The angle of incidence inside the prism will be 45. Since it is more than critical angle, the light ray will be internally reflected.
A person cannot see the object distinctly, when placed at a distance less than 100cm. What is the power of the spectacles that he should use to see clearly the objects placed at 25 cm?- a)+ 1.0 D
- b)+ 3.0 D
- c)+ 2.5 D
- d)10.0 D
Correct answer is option 'B'. Can you explain this answer?
A person cannot see the object distinctly, when placed at a distance less than 100cm. What is the power of the spectacles that he should use to see clearly the objects placed at 25 cm?
a)
+ 1.0 D
b)
+ 3.0 D
c)
+ 2.5 D
d)
10.0 D
|
|
Gaurav Kumar answered |
An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will be- a)60 cm
- b)140 cm
- c)170 cm
- d)150 cm
Correct answer is option 'B'. Can you explain this answer?
An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will be
a)
60 cm
b)
140 cm
c)
170 cm
d)
150 cm
|
|
Lavanya Menon answered |
Given that,
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100−(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5×6
D=30cm
At this instant,
Distance between mirror and object
D=100−30
D=70cm
Now, the distance between image and object is
D=70−(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100−(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5×6
D=30cm
At this instant,
Distance between mirror and object
D=100−30
D=70cm
Now, the distance between image and object is
D=70−(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.
Light rays from a point object- a)radiate in some lines
- b)radiate only to the side
- c)radiate in all directions
- d)radiate only at the front
Correct answer is option 'C'. Can you explain this answer?
Light rays from a point object
a)
radiate in some lines
b)
radiate only to the side
c)
radiate in all directions
d)
radiate only at the front
|
|
Ramesh Chand answered |
Light from each point on a luminous object travels outward in all directions in straight lines. Light travel at very high speeds, but is not instantaneously everywhere. Light from point object travels indefinitely until it collides with matter in its path to be partially absorbed and reflected.
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