All questions of Ray Optics and Optical Instruments for Airforce X Y / Indian Navy SSR Exam

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.

The light from man M₁​ hits the mirror and then reaches the man M_2​. The positions of the moving man need to be found when the light from him hits the corners of the mirror to reach the standing man.
From geometry,
AO′=L/2​+2(L/2​)
=3L/2
​Similarly, BO′=(3L/2​)+2(3L/2​)
=9L/2
Hence AB=(9L/2)-(3L/2)=3/2
Time taken to cover this distance=3L/u​
Similarly time taken to cover CD=3L/u​
Thus total time he is visible is 6L/u​
 

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). 

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.

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/f_air={(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/f_water={(1.5/1.33)-1}{(1/R₁)-(1/R₂)}
1/f_water=0.128{(1/R₁)-(1/R₂)}….2
f_water/20=0.5/0.128
f_water=78.125cm

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.

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

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

Magnification in microscope is  m=− lD/f₀​​f_e
where, l is length of tube and D is least distance for clear vision
f_0​ 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.
 

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∘

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

The reflected ray will turn 100 due to incident ray and 400 due to mirror in opposite direction 
∴ Net rotation =40−10
=300
Hence,
300 is the correct answer.

P_eq=2P_L1+2P_L2+P_M
=(2/fL1)+2(1/f_L2)-(1/f_M)
P_eq=-2(1/12)+(4/45)-(2/-15)
Peq=1/18
-1/f=1/18
F=-18cm
Here the system acts as a concave mirror.

Only Convex Mirror can form an image which is upright and reduced in size.

The ray strikes M₂ normally
so r=90^o-40^o=50^o
to the horizontal
=40 to the normal
As i=r
i=40^o

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.

The ability of the eye in terms of variation of the focal length to form sharp images on the retina is called accommodation.
 
In simpler terms, the ability of the eye to change the focus from a near distance or a far distance is called accommodation of the eye.
It is possible because of the lens which changes its shape. This makes it possible for the eye to keep focus on an object even when its distance from the eye changes.
 

Magnifying power of a compound- microscope is given by m=− (l​D​/f₀​f_e​)
where, l is length of tube
D is least distance of clear vision
f₀​ is focal length of objective
f_e​ 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.
 

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.
 

δ ACW = 180° – 60°
= 120° Anticlockwise
δ CW = 180° + 60°
= 240° Clockwise

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.

Paraxial rays are ones which originate from a point source and make a small angle with the principal axis. After reflection, they give a point image, either real or virtual.
But if rays from point source that are far from the mirror axis, gives a blurred image, an effect called spherical aberration.

Lateral inversion-The phenomenon of left side appearing right side and right side appearing left side on reflection in a plane mirror is called lateral inversion.
Letters that do not show lateral inversion are,
A, H, I, M, O, T, U, V, W, X, Y
Hence, HOX will not show lateral inversion. 

A concave mirror can form both real and virtual images depending upon the distance between object and mirror.

A convex mirror can only form virtual, erect, and diminished images; therefore, it is true that it can never form a real image of a real object. 

-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 , image of it is formed between centre of curvature and focus which is diminished,real and inverted,so real image of real object can be formed by concave mirror.
-When the object is virtual, the image formed is real for the concave surface as shown in figure.

Let ray strikes the vertical mirror by angle of incidence θ. Then by joining the perpendiculars of two mirrors it is clear from the diagram that angle of incidence at the horizontal mirror is (90−θ).
 
∴ reflected ray from the horizontal mirror makes angle θ with the horizontal mirror also incident ray at vertical mirror makes angle θ with horizontal axis. 
∴ reflected ray is parallel to the original ray.
This is true for two perpendicular mirrors fixed at any positions.

The number of images between two plane mirror inclined at an angle θ when the object is placed symmetrically between the mirrors is given as
n=(360o​/ θ) −1
Given, 
θ=60o
∴n=(360^o​/60)−1=6−1=5

α (Radian)=10/2000=1/200rad
α=h/F=h/1.2
h= α x 1.2
h=(1/200) x1.2 cm
=6mm

Solution: Let the radius of curvature of the convex surface of the plano-convex lens be RRR. The focal length of a concave mirror is related to the radius of curvature by f=R2f = \frac{R}{2}f=2R​.

By using the relations for combined focal length and refractive index of the plano-convex lens, and applying these boundary conditions, the refractive index nnn of the material of the lens is determined to be 1.5.

The Critical Angle
a light ray is in the more dense medium and approaching the less dense medium.
The angle of incidence for the light ray is greater than the so-called critical angle.
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. For any angle of incidence greater than the critical angle, light will undergo total internal reflection.

A reflecting telescope (also called a reflector) is a telescope that uses a single or a combination of curved mirrors that reflect light and form an image. The reflecting telescope was invented in the 17th century, by Isaac Newton, as an alternative to the refracting telescope which, at that time, was a design that suffered from severe chromatic aberration. Although reflecting telescopes produce other types of optical aberrations, it is a design that allows for very large diameter objectives.

Astronomical (reflecting) telescopes. In a reflecting telescope, instead of a convex objective lens, a concave mirror is used to collect parallel rays from the object and form an image at the focal point. Then the convex eyepiece lens is used to magnify this image for the viewer.