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All questions of Light Refraction and Lenses for Grade 8 Exam
Speed of light in vacuum is:- a)3 x 108 m/s
- b)3 x 109 m/s
- c)2 x 109 m/s
- d)2 x 1011 m/s
Correct answer is option 'A'. Can you explain this answer?
Speed of light in vacuum is:
a)
3 x 108 m/s
b)
3 x 109 m/s
c)
2 x 109 m/s
d)
2 x 1011 m/s
|
Krishna Iyer answered |
The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. It's exact value is defined as 299 792 458 metres per second ( or approximately, 3*108 m/s)
Transparent medium is one :
- a)Which allows light to pass through
- b)Which absorbs most of the light
- c)Which do not allows light to pass through
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Transparent medium is one :
a)
Which allows light to pass through
b)
Which absorbs most of the light
c)
Which do not allows light to pass through
d)
None of these
|
Imk Pathshala answered |
A transparent medium is a material that partially allows light to pass through.
When light encounters a transparent medium, it can penetrate the material and transmit through it.
Examples of transparent mediums include glass, water, and air.
Transparent mediums are essential for various applications such as optics, windows, and lenses.
A lens has a power of +0.5 D. It is - a)a concave lens of focal length 5 m
- b)a convex lens of focal length 5 cm
- c)a convex lens of focal length 2 m
- d)a concave lens of focal length 2 m
Correct answer is option 'C'. Can you explain this answer?
A lens has a power of +0.5 D. It is
a)
a concave lens of focal length 5 m
b)
a convex lens of focal length 5 cm
c)
a convex lens of focal length 2 m
d)
a concave lens of focal length 2 m
|
Sanya jain answered |
Given, power of the lens = 0.5 D.
We know that the power of the lens is given by the formula:
Power (P) = 1/f, where f is the focal length of the lens.
Therefore, we can write:
0.5 D = 1/f
Solving for f, we get:
f = 1/0.5 D
f = 2 m
Hence, the focal length of the lens is 2 m.
Therefore, the correct option is (c) a convex lens of focal length 2 m.
Explanation:
A convex lens has a positive power and can converge the light rays to a point. The focal length of a convex lens is positive. When the power of the lens is given, we can find the focal length of the lens using the formula P = 1/f. Here, the power is given as 0.5 D. On substituting the values, we get the focal length as 2 m. Hence, the lens is a convex lens of focal length 2 m.
We know that the power of the lens is given by the formula:
Power (P) = 1/f, where f is the focal length of the lens.
Therefore, we can write:
0.5 D = 1/f
Solving for f, we get:
f = 1/0.5 D
f = 2 m
Hence, the focal length of the lens is 2 m.
Therefore, the correct option is (c) a convex lens of focal length 2 m.
Explanation:
A convex lens has a positive power and can converge the light rays to a point. The focal length of a convex lens is positive. When the power of the lens is given, we can find the focal length of the lens using the formula P = 1/f. Here, the power is given as 0.5 D. On substituting the values, we get the focal length as 2 m. Hence, the lens is a convex lens of focal length 2 m.
The muscles of the iris control the- a)focal length of the eye-lens
- b)opening of the pupil
- c)shape of the crystalline lens
- d)optic nerve
Correct answer is option 'B'. Can you explain this answer?
The muscles of the iris control the
a)
focal length of the eye-lens
b)
opening of the pupil
c)
shape of the crystalline lens
d)
optic nerve
|
Nk Classes answered |
The Laws of Reflection for Mirrors
- The laws of reflection apply to all mirrors, regardless of their shape.
- These laws state that the angle of incidence is equal to the angle of reflection.
- This means that the light rays that strike the mirror are reflected at the same angle.
- Whether the mirror is concave, convex, or plane, these laws will always hold true.
- Therefore, option B is correct as it includes all mirrors, not just a specific type.
- The laws of reflection apply to all mirrors, regardless of their shape.
- These laws state that the angle of incidence is equal to the angle of reflection.
- This means that the light rays that strike the mirror are reflected at the same angle.
- Whether the mirror is concave, convex, or plane, these laws will always hold true.
- Therefore, option B is correct as it includes all mirrors, not just a specific type.
An object of size 2.0 cm is placed perpendicular to the principal axis of a concave mirror. The distanceof the object from the mirror equals the radius of curvature. The size of the image will be - a)0.5 cm
- b)1.5 cm
- c)1.0cm
- d)2.0 cm
Correct answer is option 'D'. Can you explain this answer?
An object of size 2.0 cm is placed perpendicular to the principal axis of a concave mirror. The distance
of the object from the mirror equals the radius of curvature. The size of the image will be
a)
0.5 cm
b)
1.5 cm
c)
1.0cm
d)
2.0 cm
|
Ishan Choudhury answered |
The size of the image will be 2.0 cm.
V=U(given)
m=-v/u
m= - 1
m=h`/h
-1=h`/2
h`=-2
h=2cm
V=U(given)
m=-v/u
m= - 1
m=h`/h
-1=h`/2
h`=-2
h=2cm
Can you explain the answer of this question below:Light travels faster in substance X as compared to substance Y, then refractive index is greater for
- A:
Can’t tell with the information provided
- B:
Substance-Y
- C:
Refractive index for both is same
- D:
Substance-X
The answer is b.
Light travels faster in substance X as compared to substance Y, then refractive index is greater for
Can’t tell with the information provided
Substance-Y
Refractive index for both is same
Substance-X
|
Gaurav Kumar answered |
As the refractive index increases, the velocity of light decreases.
The angle of incidence is the angle between- a)the incident ray and the surface of the mirror
- b)the reflected ray and the surface of the mirror
- c)the normal to the surface and the incident ray
- d)the normal to the surface and the reflected ray
Correct answer is option 'C'. Can you explain this answer?
The angle of incidence is the angle between
a)
the incident ray and the surface of the mirror
b)
the reflected ray and the surface of the mirror
c)
the normal to the surface and the incident ray
d)
the normal to the surface and the reflected ray
|
|
Hitakshi 💞💞 answered |
The angle between the incident ray and the normal is called as angle of incidence and it is generally denoted by i. The angle between the normal and the reflected ray is called as angle of reflection and it is generally denoted by r.
An object is placed before a concave lens. The image formed- a)is always erect
- b)may be erect or inverted
- c)is always inverted
- d)is always real
Correct answer is option 'A'. Can you explain this answer?
An object is placed before a concave lens. The image formed
a)
is always erect
b)
may be erect or inverted
c)
is always inverted
d)
is always real
|
Headway Institute answered |
Because concave lens always produce an
Image which is always erect, diminished and virtual.
Image which is always erect, diminished and virtual.
Refractive index of glass w.r.t. air is 3/2. What is the refractie index of air w.r.t glass ?- a)2/3
- b)1
- c)Zero
- d)(3/2)2
Correct answer is option 'A'. Can you explain this answer?
Refractive index of glass w.r.t. air is 3/2. What is the refractie index of air w.r.t glass ?
a)
2/3
b)
1
c)
Zero
d)
(3/2)2
|
Zara Khan answered |
If refractive index of glass w.r.t air is 3/2 , then refractive index of air w. r. t glass will be it's reciprocal ie. ⅔ ..
An object is placed before a convex lens. The image formed- a)is always real
- b)may be real or virtual
- c)is always virtual
- d)is always erect
Correct answer is option 'B'. Can you explain this answer?
An object is placed before a convex lens. The image formed
a)
is always real
b)
may be real or virtual
c)
is always virtual
d)
is always erect
|
EduRev Class 10 answered |
Convex Lens: A convex lens is thicker at the center and thinner at the edges. When light rays pass through a convex lens, they converge at a point called the focal point.
Image Formation: When an object is placed before a convex lens, the light rays from the object refract through the lens and form an image on the other side. The image formation by a convex lens depends on the distance of the object from the lens.
Real or Virtual Image: The image formed by a convex lens can be real or virtual, depending on the position of the object relative to the lens.
Real Image: A real image is formed when the light rays actually converge at a point after passing through the lens. This real image can be projected onto a screen and is always inverted.
- Virtual Image: A virtual image is formed when the light rays appear to converge at a point on the same side of the lens as the object. This virtual image cannot be projected onto a screen and is always upright.
- Conclusion: Therefore, when an object is placed before a convex lens, the image formed may be real or virtual, depending on the position of the object relative to the lens.
A point object is placed on the principal axis of a spherical mirror. The object-distance u is - a)definitely negative
- b)definitely positive
- c)positive if the object is to the left of the centre of curvature
- d)positive if the object is to the right of the centre of curvature
Correct answer is option 'A'. Can you explain this answer?
A point object is placed on the principal axis of a spherical mirror. The object-distance u is
a)
definitely negative
b)
definitely positive
c)
positive if the object is to the left of the centre of curvature
d)
positive if the object is to the right of the centre of curvature
|
Ritu Saxena answered |
Option ( a) is the correct answer. As the object is always placed on the left side of the mirror and according to the sign convention, it has negative value for 'so axis.
Therefore, spherical mirrors have only one reflecting surface and it will be negative only.
The speed of light in glass is 1.5 x 108m/s. Then refractive index of glass is
- a)3
- b)1
- c)4
- d)2
Correct answer is option 'D'. Can you explain this answer?
The speed of light in glass is 1.5 x 108m/s. Then refractive index of glass is
a)
3
b)
1
c)
4
d)
2
|
Palak Chatterjee answered |
Refractive index of glass
Refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v):
n = c/v
Given that the speed of light in glass (v) is 1.5 x 108 m/s, we can calculate the refractive index of glass as follows:
n = c/v = 3 x 108 m/s / 1.5 x 108 m/s = 2
Therefore, the refractive index of glass is 2.
Option D is the correct answer.
Refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v):
n = c/v
Given that the speed of light in glass (v) is 1.5 x 108 m/s, we can calculate the refractive index of glass as follows:
n = c/v = 3 x 108 m/s / 1.5 x 108 m/s = 2
Therefore, the refractive index of glass is 2.
Option D is the correct answer.
An object is placed at the centre of curvature of a concave mirror. The distance between its imageand the pole is - a)equal to f
- b)between f and 2f
- c)equal to 2f
- d)greater than 2f
Correct answer is option 'C'. Can you explain this answer?
An object is placed at the centre of curvature of a concave mirror. The distance between its image
and the pole is
a)
equal to f
b)
between f and 2f
c)
equal to 2f
d)
greater than 2f
|
Ishaani mehta answered |
Explanation:
The given situation can be represented as follows:
Here, C is the centre of curvature of the concave mirror, F is the focus, and P is the pole.
Now, let us consider the path of light rays from the object to the mirror and then to its image.
- The light rays from the object are parallel to the principal axis and fall on the mirror.
- At the point of incidence, the light rays are reflected and converge towards the focus F.
- However, since the object is placed at the centre of curvature C, the reflected rays pass through F and become parallel to the principal axis.
- These parallel rays then converge at the position of the image I.
Therefore, we can see that the image is formed at a distance equal to twice the focal length from the mirror.
- From the mirror formula:
1/f = 1/v + 1/u
where f is the focal length, v is the image distance, and u is the object distance.
- In this case, since the object is placed at C, u = -2f (negative sign indicates that it is on the opposite side of the mirror).
- We want to find v, the distance between the image and the pole.
- Substituting the given values, we get:
1/f = 1/v - 1/2f
Multiplying both sides by vf2, we get:
2f = v + f
Therefore, v = 2f - f = f.
Hence, the distance between the image and the pole is equal to the focal length of the mirror, which is option C.
The given situation can be represented as follows:
Here, C is the centre of curvature of the concave mirror, F is the focus, and P is the pole.
Now, let us consider the path of light rays from the object to the mirror and then to its image.
- The light rays from the object are parallel to the principal axis and fall on the mirror.
- At the point of incidence, the light rays are reflected and converge towards the focus F.
- However, since the object is placed at the centre of curvature C, the reflected rays pass through F and become parallel to the principal axis.
- These parallel rays then converge at the position of the image I.
Therefore, we can see that the image is formed at a distance equal to twice the focal length from the mirror.
- From the mirror formula:
1/f = 1/v + 1/u
where f is the focal length, v is the image distance, and u is the object distance.
- In this case, since the object is placed at C, u = -2f (negative sign indicates that it is on the opposite side of the mirror).
- We want to find v, the distance between the image and the pole.
- Substituting the given values, we get:
1/f = 1/v - 1/2f
Multiplying both sides by vf2, we get:
2f = v + f
Therefore, v = 2f - f = f.
Hence, the distance between the image and the pole is equal to the focal length of the mirror, which is option C.
A ray of light is incident on a concave mirror. If it is parallel to the principal axis, the reflected ray will- a)pass through the focus
- b)pass through the centre of curvature
- c)pass through the pole
- d)retrace its path
Correct answer is option 'A'. Can you explain this answer?
A ray of light is incident on a concave mirror. If it is parallel to the principal axis, the reflected ray will
a)
pass through the focus
b)
pass through the centre of curvature
c)
pass through the pole
d)
retrace its path
|
|
Diana Nair answered |
A ray of light which is parallel to the principle axis after reflection will pass through the focus of a concave mirror.
Indicate the only correct statement.- a)The image formed by a convex mirror can be taken on the screen.
- b)A convex mirror can produce a parallel beam of light from a point source.
- c)The image of'an object placed at the focus of a convex mirror will be formed at infinity.
- d)A concave mirror can never form a diminished virtual image.
Correct answer is option 'D'. Can you explain this answer?
Indicate the only correct statement.
a)
The image formed by a convex mirror can be taken on the screen.
b)
A convex mirror can produce a parallel beam of light from a point source.
c)
The image of'an object placed at the focus of a convex mirror will be formed at infinity.
d)
A concave mirror can never form a diminished virtual image.
|
|
EduRev Class 10 answered |
Option D: "A concave mirror can never form a diminished virtual image."
Correct. A concave mirror can form a diminished real image but never a diminished virtual image. Virtual images formed by concave mirrors are always enlarged.
The angle of reflection is the angle between - a)the incident ray and the surface of the mirror
- b)the reflected ray and the surface of the mirror
- c)the normal to the surface and the incident ray
- d)the normal to the surface and the reflected ray
Correct answer is option 'D'. Can you explain this answer?
The angle of reflection is the angle between
a)
the incident ray and the surface of the mirror
b)
the reflected ray and the surface of the mirror
c)
the normal to the surface and the incident ray
d)
the normal to the surface and the reflected ray
|
|
Krishna Iyer answered |
Angle of reflection is the angle between a reflected ray and the normal drawn at the point of incidence to a reflecting surface.
Which of the following can be used to form a virtual image of an object?
I. convex lens
II. concave lens
III. concave mirror- a)II only
- b)II and III only
- c)I and III only
- d)I, II and III
Correct answer is option 'D'. Can you explain this answer?
Which of the following can be used to form a virtual image of an object?
I. convex lens
II. concave lens
III. concave mirror
I. convex lens
II. concave lens
III. concave mirror
a)
II only
b)
II and III only
c)
I and III only
d)
I, II and III
|
Adidev Nair answered |
A concave lens forms
real and inverted image
. When the
object
is placed very close to the
lens
, the
image
formed is
virtual
, erect and magnified.
A
concave lens
always
forms
erect,
virtual
and smaller
image
than the
object
.
Concave
mirrors
can produce
both real and
virtual images
; they
can
be upright (if
virtual
) or inverted (if real); they
can
be behind the
mirror
(if
virtual
) or in front of the
mirror
(if real); they
can
also be enlarged, reduced, or the same size as
object
.
The speed of light in vacuum is 3.0 × 108 m/s. If the refractive index of a transparent liquid is 4/3, then the speed of light in the liquid is- a)2.25 × 108 m/s
- b)3 x 108 m/s
- c)4 × 108 m/s
- d)4.33 × 108 m/s
Correct answer is option 'A'. Can you explain this answer?
The speed of light in vacuum is 3.0 × 108 m/s. If the refractive index of a transparent liquid is 4/3, then the speed of light in the liquid is
a)
2.25 × 108 m/s
b)
3 x 108 m/s
c)
4 × 108 m/s
d)
4.33 × 108 m/s
|
Gunjan Lakhani answered |
Refractive index of water :-
Speed of light in vacuum / speed of light in water
4/3 = (3*108) /x
x = 2.25 * 108 m/s
Speed of light in vacuum / speed of light in water
4/3 = (3*108) /x
x = 2.25 * 108 m/s
A concave mirror of focal length 10 cm produces an image five times as large as the object. If the image is in front of the mirror, the distance of the object from the mirror will be
- a) 10 cm
- b) 12 cm
- c) 16 cm
- d) 20 cm
Correct answer is option 'B'. Can you explain this answer?
A concave mirror of focal length 10 cm produces an image five times as large as the object. If the image is in front of the mirror, the distance of the object from the mirror will be
a)
10 cmb)
12 cmc)
16 cmd)
20 cm
|
Amit Kumar answered |
Here, f = -10 cmcm
m = -5 ( image is 5 times larger than the object)
⇒ -v/ u = -5
⇒ v = 5u
By using mirror formula,
1/f = 1/u + 1/v
⇒ 1/-10 = 1/u +1/5u
⇒ 1/-10 = 6/5u
⇒ 5u = 6*(-10) cm
⇒ u = -60/5 cm
⇒ u = -12 cm
So, the distance of object is 12 cm from the mirror.
m = -5 ( image is 5 times larger than the object)
⇒ -v/ u = -5
⇒ v = 5u
By using mirror formula,
1/f = 1/u + 1/v
⇒ 1/-10 = 1/u +1/5u
⇒ 1/-10 = 6/5u
⇒ 5u = 6*(-10) cm
⇒ u = -60/5 cm
⇒ u = -12 cm
So, the distance of object is 12 cm from the mirror.
Can you explain the answer of this question below:MCQ (Multiple Choice Questions) or Practice Quiz with solutions of Chapter - "Ray Optics" of Class 10 Science, the questions are available for practice
Q. A mirror forms a virtual image of a real object.
- A:
It must be a convex mirror.
- B:
It must be a concave mirror.
- C:
It must be a plane mirror.
- D:
It may be any of the mirrors mentioned above.
The answer is D.
MCQ (Multiple Choice Questions) or Practice Quiz with solutions of Chapter - "Ray Optics" of Class 10 Science, the questions are available for practice
Q. A mirror forms a virtual image of a real object.
It must be a convex mirror.
It must be a concave mirror.
It must be a plane mirror.
It may be any of the mirrors mentioned above.
|
|
Krishna Iyer answered |
The image in a plane mirror forms virtual and erect image and appears to be as far behind the mirror as the object is in front of the mirror. A diverging lens or a convex mirror forms a virtual image.A virtual image is produced by a concave mirror when the object is placed inside the focal length of the mirror.
To get an image larger than the object, one can use - a)a convex mirror but not a concave mirror
- b)a concave mirror but not a convex mirror
- c)either a convex mirror or a concave mirror
- d)a plane mirror
Correct answer is option 'B'. Can you explain this answer?
To get an image larger than the object, one can use
a)
a convex mirror but not a concave mirror
b)
a concave mirror but not a convex mirror
c)
either a convex mirror or a concave mirror
d)
a plane mirror
|
|
Rohit Sharma answered |
A convex mirror never forms an image larger than the original object, whereas a concave mirror can form an image larger than the object between the centre of curvature and the point of focus.
If an incident ray passes through the centre of curvature of a spherical mirror, the reflected ray will - a)pass through the pole
- b)pass through the focus
- c)retrace its path
- d)be parallel to the principal axis
Correct answer is option 'C'. Can you explain this answer?
If an incident ray passes through the centre of curvature of a spherical mirror, the reflected ray will
a)
pass through the pole
b)
pass through the focus
c)
retrace its path
d)
be parallel to the principal axis
|
Naina Sharma answered |
An incident ray which passes through the centre of curvature C of the mirror is reflected back along it's own path ( since it is normally incident on the mirror) .
A convex lens forms a virtual image when an object is placed at a distance of 18 cm from it. The focal length must be- a)greater than 36 cm
- b)greater than 18 cm
- c)less than 36 cm
- d)less than 18 cm
Correct answer is option 'B'. Can you explain this answer?
A convex lens forms a virtual image when an object is placed at a distance of 18 cm from it. The focal length must be
a)
greater than 36 cm
b)
greater than 18 cm
c)
less than 36 cm
d)
less than 18 cm
|
|
Amit Kumar answered |
A virtual image of an object is formed by a convex lens only when the object is placed between the lens and the focus. Therefore, the focal length of the lens should be greater than the distance of the object from the lens.
A ray of light goes from a medium of refractive index μ1 to a medium of refractive index μ2. Theangle of incidence is i and the angle of refraction is r. Then, sin i/sin r is equal to- a)μ1
- b)μ2
- c)μ1/μ2
- d)μ2/μ1
Correct answer is option 'D'. Can you explain this answer?
A ray of light goes from a medium of refractive index μ1 to a medium of refractive index μ2. Theangle of incidence is i and the angle of refraction is r. Then, sin i/sin r is equal to
a)
μ1
b)
μ2
c)
μ1/μ2
d)
μ2/μ1
|
|
Amit Sharma answered |
Refractive index, also called index of refraction, measure of the bending of a ray of light when passing from one medium into another. As per Snell's law and the given information ;
A convex lens- a)is thicker at the middle than at the edges
- b)is thicker at the edges than at the middle
- c)has uniform thickness everywhere
- d)is called a diverging lens
Correct answer is option 'A'. Can you explain this answer?
A convex lens
a)
is thicker at the middle than at the edges
b)
is thicker at the edges than at the middle
c)
has uniform thickness everywhere
d)
is called a diverging lens
|
Kamna Science Academy answered |
Explanation:
Convex lens A: is thicker at the middle than at the edges
Convex lenses are thicker at the center and thinner towards the edges.
This shape causes light rays passing through the lens to converge or come together at a focal point.
This property makes convex lenses useful for focusing light in applications such as cameras, eyeglasses, and projectors.
The magnification of an object placed 10 cm from a convex mirror of radius of curvature 20 cmwill be - a)0.2
- b)0.5
- c)1
- d)infinity
Correct answer is option 'B'. Can you explain this answer?
The magnification of an object placed 10 cm from a convex mirror of radius of curvature 20 cm
will be
a)
0.2
b)
0.5
c)
1
d)
infinity
|
|
Rohan Kapoor answered |
0.5 is the magnification
it came because....
u= -10cm
f=10 cm
v=5cm {1/f=1/v+1/u}
magnification=-v/u
m=5/10
magnification=0.5
A ray of light travelling in air falls obliquely on the surface of a calm pond. It will - a)go into the water without deviating from its path
- b)deviate away from the normal
- c)deviate towards the normal
- d)turn back on its original path
Correct answer is option 'C'. Can you explain this answer?
A ray of light travelling in air falls obliquely on the surface of a calm pond. It will
a)
go into the water without deviating from its path
b)
deviate away from the normal
c)
deviate towards the normal
d)
turn back on its original path
|
|
Sameer nanda answered |
Explanation:
When a ray of light travelling in air falls obliquely on the surface of a calm pond, it undergoes refraction. The incident ray, refracted ray and the normal to the surface of the pond all lie in the same plane.
Refraction of Light:
The bending of light as it passes from one medium to another is called refraction. When light passes from one medium to another, its speed and direction change. The speed of light is different in different media.
Law of Refraction:
The law of refraction, also known as Snell's law, states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media. Mathematically, it can be written as:
sin i / sin r = constant
where i is the angle of incidence, r is the angle of refraction, and the constant is called the refractive index.
Deviation of Light:
The deviation of light depends on the angle of incidence and the refractive index of the medium. When a ray of light passes from a rarer medium to a denser medium, it bends towards the normal. When a ray of light passes from a denser medium to a rarer medium, it bends away from the normal.
Answer:
When a ray of light travelling in air falls obliquely on the surface of a calm pond, it passes from a rarer medium (air) to a denser medium (water). Therefore, the ray of light will deviate towards the normal. Hence, option (c) is the correct answer.
When a ray of light travelling in air falls obliquely on the surface of a calm pond, it undergoes refraction. The incident ray, refracted ray and the normal to the surface of the pond all lie in the same plane.
Refraction of Light:
The bending of light as it passes from one medium to another is called refraction. When light passes from one medium to another, its speed and direction change. The speed of light is different in different media.
Law of Refraction:
The law of refraction, also known as Snell's law, states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media. Mathematically, it can be written as:
sin i / sin r = constant
where i is the angle of incidence, r is the angle of refraction, and the constant is called the refractive index.
Deviation of Light:
The deviation of light depends on the angle of incidence and the refractive index of the medium. When a ray of light passes from a rarer medium to a denser medium, it bends towards the normal. When a ray of light passes from a denser medium to a rarer medium, it bends away from the normal.
Answer:
When a ray of light travelling in air falls obliquely on the surface of a calm pond, it passes from a rarer medium (air) to a denser medium (water). Therefore, the ray of light will deviate towards the normal. Hence, option (c) is the correct answer.
A deviation in the path of a ray of light can be produced- a)by a glass prism as well as a rectangular glass slab
- b)by a rectangular glass slab but not by a glass prism
- c)by a glass prism but not by a rectangular glass slab
- d)neither by a glass prism nor by a rectangular glass slab
Correct answer is option 'A'. Can you explain this answer?
A deviation in the path of a ray of light can be produced
a)
by a glass prism as well as a rectangular glass slab
b)
by a rectangular glass slab but not by a glass prism
c)
by a glass prism but not by a rectangular glass slab
d)
neither by a glass prism nor by a rectangular glass slab
|
Pooja Shah answered |
The bending of light as it passes from one medium to another is called refraction. The angle and wavelength at which the light enters a substance and the density of that substance determine how much the light is refracted. The bending occurs because light travels more slowly in a denser medium.
Hence, by a glass prism as well as a rectangular glass slab
What will happen when a ray enters from glass to air medium ?
- a)Totally reflected from air surface.
- b)It bends away from normal.
- c)Follows its original path.
- d)It bends towards the normal.
Correct answer is option 'B'. Can you explain this answer?
What will happen when a ray enters from glass to air medium ?
a)
Totally reflected from air surface.
b)
It bends away from normal.
c)
Follows its original path.
d)
It bends towards the normal.
|
|
Alok Saini answered |
When a ray of light passes from a less dense material into denser material it is bent away from the surface between the two materials. This means that in this situation the angle of refraction is always less than the angle of incidence.
To form an image twice the size of the object, using a convex lens of focal length 20 cm, the object distance must be –- a)< 20 cm
- b)> 20 cm
- c)> 20 cm and between 20 cm and 40 cm
- d)Cannot form image
Correct answer is option 'C'. Can you explain this answer?
To form an image twice the size of the object, using a convex lens of focal length 20 cm, the object distance must be –
a)
< 20 cm
b)
> 20 cm
c)
> 20 cm and between 20 cm and 40 cm
d)
Cannot form image
|
Kds Coaching answered |
Given, f = 20 cm, h1 = 2h0
⇒ h1/ho = m = 2
v / u = 2 => v = 2u
1/ v - 1/ u = 1/ f ( lens formula)
⇒ 1/2u - 1/-u = 1/ 20
⇒ 3/2u = 1/20
⇒ u = 30 cm
So, option (c) is the correct answer.
A dentist uses a small concave mirror with a focal length of 1.6 cm to observe a cavity in a patient's tooth. The mirror is held 8 mm (0.8 cm) from the cavity. What is the magnification produced by the mirror?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
A dentist uses a small concave mirror with a focal length of 1.6 cm to observe a cavity in a patient's tooth. The mirror is held 8 mm (0.8 cm) from the cavity. What is the magnification produced by the mirror?
a)
1
b)
2
c)
3
d)
4
|
|
Nk Classes answered |
Given:
- Focal length (f) = 1.6 cm
(Since it's a concave mirror, focal length is negative, so f = -1.6 cm) - Object distance (u) = -0.8 cm
(Object is real, so u is negative)
Mirror Formula:
1/f = 1/v + 1/u
Substitute the values:
1/(-1.6) = 1/v + 1/(-0.8)
Now solve step-by-step:
1/(-1.6) = -0.625
1/(-0.8) = -1.25
1/(-0.8) = -1.25
So,
-0.625 = 1/v - 1.25
Now move -1.25 to the left side:
1/v = -0.625 + 1.25
1/v = 0.625
1/v = 0.625
Now take reciprocal:
v = 1 / 0.625 = 1.6 cm
Now, find magnification (m):
m = -v / u
m = -1.6 / (-0.8)
m = 2
m = -1.6 / (-0.8)
m = 2
Final Answer: b) 2
In a museum, a child walks towards a large concave mirror. Which of the following describes what happens to the image?- a)his real, erect image goes on decreasing in size.
- b)his virtual, erect image goes on increasing in size.
- c)his real, inverted image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified.
- d)his real, erect image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified.
Correct answer is option 'C'. Can you explain this answer?
In a museum, a child walks towards a large concave mirror. Which of the following describes what happens to the image?
a)
his real, erect image goes on decreasing in size.
b)
his virtual, erect image goes on increasing in size.
c)
his real, inverted image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified.
d)
his real, erect image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified.
|
Raghav Choudhury answered |
Understanding Concave Mirrors
Concave mirrors are unique because they can create both real and virtual images depending on the distance of the object from the mirror.
Image Formation by Concave Mirrors
- When an object is placed beyond the center of curvature (C), the mirror produces a real, inverted image that is smaller than the object.
- As the object moves closer to the mirror, specifically between the focus (F) and C, the image remains real and inverted but gradually increases in size.
- When the object moves between the focal point and the mirror, the mirror starts creating a virtual image.
Scenario with the Child Approaching the Mirror
- Initially, as the child approaches the concave mirror, they are positioned outside the focus.
- Their real and inverted image appears smaller as they get closer, but it remains real until they cross the focal point.
Key Transition Point
- Once the child crosses the focal point, the nature of the image changes:
- The image becomes virtual, meaning it cannot be projected onto a screen.
- The image also becomes erect and magnified, appearing larger than the actual size of the child.
Conclusion
Therefore, the correct answer is option 'C': "his real, inverted image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified." This accurately describes the transition of the image from real to virtual as the child approaches the concave mirror.
Concave mirrors are unique because they can create both real and virtual images depending on the distance of the object from the mirror.
Image Formation by Concave Mirrors
- When an object is placed beyond the center of curvature (C), the mirror produces a real, inverted image that is smaller than the object.
- As the object moves closer to the mirror, specifically between the focus (F) and C, the image remains real and inverted but gradually increases in size.
- When the object moves between the focal point and the mirror, the mirror starts creating a virtual image.
Scenario with the Child Approaching the Mirror
- Initially, as the child approaches the concave mirror, they are positioned outside the focus.
- Their real and inverted image appears smaller as they get closer, but it remains real until they cross the focal point.
Key Transition Point
- Once the child crosses the focal point, the nature of the image changes:
- The image becomes virtual, meaning it cannot be projected onto a screen.
- The image also becomes erect and magnified, appearing larger than the actual size of the child.
Conclusion
Therefore, the correct answer is option 'C': "his real, inverted image goes on diminishing in size and suddenly it becomes virtual, erect, and magnified." This accurately describes the transition of the image from real to virtual as the child approaches the concave mirror.
When white light passes through a prism, it splits into its component colours. This phenomenon is called- a)spectrum
- b)reflection
- c)refraction
- d)dispersion
Correct answer is option 'D'. Can you explain this answer?
When white light passes through a prism, it splits into its component colours. This phenomenon is called
a)
spectrum
b)
reflection
c)
refraction
d)
dispersion
|
|
Rajiv Gupta answered |
The dispersion of light is the phenomenon of splitting of a beam of white light into its seven constituent colours when passed through a transparent medium. It was discovered by Isaac Newton in 1666. Newton discovered that light is made up of seven different colours. He passed a beam of sunlight through a glass prism. The glass prism split the light into a band of seven colours on his wall. He called this band of colours the ‘spectrum’. Thus the spectrum is a band of seven colours which is obtained by splitting of white light by a glass prism. The order of colours from the lower end of spectrum is violet (V), indigo (I), blue (B), green (G), yellow (Y), orange (O), and red (R). The sequence of the 7 colours so obtained in a spectrum can be remembered by using the acronym ‘VIBGYOR’.
The magnification m of an image formed by a spherical mirror is negative. It means, the image is- a)smaller than the object
- b)larger than the object
- c)erect
- d)inverted
Correct answer is option 'D'. Can you explain this answer?
The magnification m of an image formed by a spherical mirror is negative. It means, the image is
a)
smaller than the object
b)
larger than the object
c)
erect
d)
inverted
|
Sreemoyee Patel answered |
For spherical mirrors,
m = - v/u
‘m’ is negative when both ‘v’ and ‘u’ have negative values. That is both object and image are on the same side.
So, the correct option is d. The image is inverted, since, it is real.
1. Which of the following statements is true regarding reflection of light?- a)The angle of incidence is always greater than the angle of reflection
- b)The angle of incidence is always equal to the angle of reflection
- c)The angle of incidence is always smaller than the angle of reflection
- d)The angle of incidence and angle of reflection are not related
Correct answer is option 'B'. Can you explain this answer?
1. Which of the following statements is true regarding reflection of light?
a)
The angle of incidence is always greater than the angle of reflection
b)
The angle of incidence is always equal to the angle of reflection
c)
The angle of incidence is always smaller than the angle of reflection
d)
The angle of incidence and angle of reflection are not related
|
|
Kamna Science Academy answered |
- The correct answer is B: The angle of incidence is always equal to the angle of reflection.
- This principle is known as the Law of Reflection.
- When light hits a smooth surface, it reflects in such a way that:
- The angle at which the light ray hits the surface (angle of incidence) is equal to the angle at which it leaves (angle of reflection).
- These angles are measured from the normal, an imaginary line perpendicular to the surface at the point of incidence.
- This principle is known as the Law of Reflection.
- When light hits a smooth surface, it reflects in such a way that:
- The angle at which the light ray hits the surface (angle of incidence) is equal to the angle at which it leaves (angle of reflection).
- These angles are measured from the normal, an imaginary line perpendicular to the surface at the point of incidence.
Which of the following conditions are necessary for total internal reflection to take place at the boundary of two optical media?I. Light is passing from optically denser medium to optically rarer medium.
II. Light is passing from optically rarer medium to optically denser medium.
III. Angle of incidence is greater than the critical angle.
IV. Angle of incidence is less than the critical angle.- a)I and III only
- b)II and IV only
- c)III and IV only
- d)I and IV only
Correct answer is option 'A'. Can you explain this answer?
Which of the following conditions are necessary for total internal reflection to take place at the boundary of two optical media?
I. Light is passing from optically denser medium to optically rarer medium.
II. Light is passing from optically rarer medium to optically denser medium.
III. Angle of incidence is greater than the critical angle.
IV. Angle of incidence is less than the critical angle.
II. Light is passing from optically rarer medium to optically denser medium.
III. Angle of incidence is greater than the critical angle.
IV. Angle of incidence is less than the critical angle.
a)
I and III only
b)
II and IV only
c)
III and IV only
d)
I and IV only
|
|
EduRev Class 10 answered |
Total internal reflection (TIR) occurs under two necessary conditions:
- Light must travel from an optically denser medium to an optically rarer medium (e.g., from water to air, or glass to air). This ensures that light bends away from the normal at the boundary. If light travels from rarer to denser, TIR cannot occur.
- Angle of incidence must be greater than the critical angle. The critical angle is the minimum angle of incidence in the denser medium for which the refracted ray in the rarer medium would travel along the boundary. Only when the incidence is greater than this angle does total reflection occur.
Why the other options are wrong:
- II: Light from rarer to denser medium cannot undergo TIR.
- IV: Angle of incidence less than the critical angle results in partial refraction, not total internal reflection.
Answer: I and III only.
The wavelengths corresponding to violet, yellow and red lights are λv , λy and λr respectively.- a)λv > λy > λr
- b)λv < λy < λr
- c)λy < λv < λr
- d)λy < λr < λv
Correct answer is option 'B'. Can you explain this answer?
The wavelengths corresponding to violet, yellow and red lights are λv , λy and λr respectively.
a)
λv > λy > λr
b)
λv < λy < λr
c)
λy < λv < λr
d)
λy < λr < λv
|
|
Kamna Science Academy answered |
Explanation:- because we know, the wavelength of red colour is the maximum, then comes yellow, and then Violet.
The angle of incidence for a light ray incident on a plane mirror is 30 degrees. What will be the angle of reflection?- a)30 degrees
- b)60 degrees
- c)90 degrees
- d)120 degrees
Correct answer is option 'B'. Can you explain this answer?
The angle of incidence for a light ray incident on a plane mirror is 30 degrees. What will be the angle of reflection?
a)
30 degrees
b)
60 degrees
c)
90 degrees
d)
120 degrees
|
|
Nk Classes answered |
- The angle of incidence is the angle between the incoming ray and the normal (a line perpendicular to the surface).
- According to the law of reflection, the angle of incidence is equal to the angle of reflection.
- Given that the angle of incidence is 30 degrees, the angle of reflection will also be 30 degrees.
- Therefore, the correct answer is A: 30 degrees.
- According to the law of reflection, the angle of incidence is equal to the angle of reflection.
- Given that the angle of incidence is 30 degrees, the angle of reflection will also be 30 degrees.
- Therefore, the correct answer is A: 30 degrees.
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