Test: Ray Optics & Optical Instruments - 3 - JEE MCQ

The angle subtended at the eye by the top of the tower with respect to the ground is 45 degrees. This happens when the height of the tower is equal to the distance from the eye. The eye is assumed to be very near to the mirror. (Refer below figure).
Angle subtended by AB = 90 degrees/2 = 45 degrees.
Therefore, tanθ=h/60​. That is, tan45=h/60​.So, h=60m.
Hence, the height of the tower is 60 m.

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​
 

Width of the room = 200cm (Given)
Length of plane mirror = 10cm  (Given)
Distance of the person looking into mirror = 50cm  (Given)
In first triangle -
TanФ = x/200 --- 1
In second triangle -
TanФ = 5/50 --- 2
Equating both the equations -
5/50 = x/200
x = 20cm
Thus, the height of the wall is 2x+10 = 50
The width of the wall he will be able to see = 50cm

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.
 

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.

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.

There form 3 images due to refraction on two different lenses 3 times one upper ,1 refraction from both lenses, 1 at the bottom.
now focus of Plano concave =f=R/(1−μ)
hence focus for diagram ii> f=R/2(1−μ)
and same focus for diagram iii>because again there is combination of two two Plano concave lens
so, the ratio of focus is 1:1
 

h_object²​=h_image1​×h_image2​
h_object​=√36​=6cm
magnification of image is v/u​=9/6​
v=3u​/2
in lens displacement method, u+v=d; uv=df
u+(3u/2)​=90
u=36   => v=54cm
uv=df 
f=(36×54)/90​=21.6
 

Refraction of light of different wavelength is shown in the figure.The light inside the slab clearly splits up into different colors. The emergent ray from a single white ray is a beam of different colors, parallel to the incident ray. However the emergent beam is white, because the white rays adjacent to the incident white ray above also split into different colors that add up to these wavelengths to produce a white beam of light parallel to the incident white beam.

We can either have deviation or not and same for dispersion as only μ is different and angle of prism is of our choice
option A,B,C are correct.

(μ/v)- (μ/u) = (μ₂- μ1)/R
Or, (μ/2R) + (1/2R) = μ-1/R
Or, (μ/2) + (1/2) = (μ-1)/1
Or, (Μ+1)/2= μ-1
Or, μ+1=2μ-2
 μ=3

 

From the symmetry of the fig.ray inside the sphere is parallel to the principal axis. Taking refraction at A
(μ₂/v)−(μ₁/u))=(μ₂−μ₁)/R
(μ/∞)−(1/−R)=(μ−1)/R
(1/R)=(μ−1)/R⇒μ−1=1
μ=2

From the symmetry of the fig.ray inside the sphere is parallel to the principal axis. Taking refraction at A
(μ₂/v)−(μ₁/u))=(μ₂−μ₁)/R
(μ/∞)−(1/−R)=(μ−1)/R
(1/R)=(μ−1)/R⇒μ−1=1
μ=2

a) At first the height of the liquid level is obviously dependent because the normal shift will happen according to the depth of the liquid. 
Drawing a ray diagram keeping a coin at very deep and another close to the surface.We will notice that the earlier one will have a noticeable change of position but the later one will only change a bit. 
b) The position of the slab is independent because wherever we keep it the incident ray & the emergent ray at last will be parallel (same medium keeping in mind).
 

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

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

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.

Figure shows two plane mirrors kept at angle 60°.  A light ray incidents at A making an angle of incidence α,
reflected and again incidents at B making an angle β and getting reflected .
Reflected ray from B makes angle θ with the initial incident ray.
At reflection point A, as shown in figure, we have α+x = 90° ..............(1)
At reflection point B, as shown in figure, we have β+y = 90° ..............(2)
By adding (1) and (2), after substituting x+y = 120°, we get,  α+β = 60° ...............(3)
From ΔABC, we have,  2α + 2β + θ = 180° ...................(4)
Hence By using eqn.(3), we get , θ = 180 - 120 = 60°
 

Here, given that,
v=-6.2 and u=26
we have to find,f=?
So, we know,
1/f=(1/u)+(1/v)
1/f=(1/-6.2) + (1/26)
f=-8.41cm
Hence the correct answer is option B.

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