Test: Ray Optics & Optical Instruments - 1 - JEE MCQ

Light from each point on a luminous object travels outward in all directions in straight lines. Light travels at very high speeds, but is not instantaneously everywhere. Light from a point object travels indefinitely until it collides with matter in its path to be partially absorbed and reflected.

In a convex lens,  (1/v)−(1/u)=(1/f)​
Or (1/v)​−(1/f)​=(1/u)​ where u is always negative and f is always positive.
(1/v)​=(1/f)​−(1/u)​
So, as u is increased 1/u will decrease and  (1/f)​−(1/u)​  will increase, so, 1/v​ increases or v decreases.
So, its graph will be a hyperbola

P=P1​+P2​=+2−1=+1 dioptre, lens behaves as convergent
F=1​/P=1/1​=1m=100cm

In air, all the colours of light travel with the same velocity with the same velocity, but in glass, velocities of different colours are different. Velocity of red colour is largest and velocity of violet colour is smallest. Therefore, after travelling through the glass slab, red colour will emerge first.

When the deviation is minimum, then the incidence angle and emergence angle are equal and this is possible only when the refracted ray inside the prism is parallel to the base of the prism. 
Minimum deviation is given by δ=2i−A, where A is prism angle.

Magnification of a telescope is given by:
M= θ_i/θ_o​ ​​
10= (h/d_i)/(h/d_o​) ​​
10= d_o​​/d_i​
Distance of image, d_i​=d_o​​/10

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.
 

The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light. These particles are more effective in scattering light of a shorter wavelength at the blue end than the light of the longer wavelengths at the red end. The red light has a wavelength 1.8  times greater than blue light. Thus, when sunlight passes through the atmosphere, the fine particles in the air scatter the blue color more strongly than red color. The scattered blue light enters our eyes and we see the blue color of the clear sky.

Magnifying power of a telescope is f₀/f_e​ ​​, so as  1/​​f_e increases, magnifying power increases.
Hence the correct answer is option D.

L₁L₂=40cm and O I=100cm
Therefore, x+40+x=100 or x=30cm
For lens at L1, we have
U=-30cm and v=+70
Thus,
(1/v)-(1/u)=1/f
Or, 1/f=(1/70) - (1/-30)
Or, f=+21cm
 

The ability of the eye lens to adjust its focal length is called power of accommodation. This is done by the ciliary muscles by changing the focal length of the eye lens.

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.

3. 1.5

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​.

1. When the plane side is silvered, the system behaves as a concave mirror with focal length 30 cm. The effective focal length fefff_{\text{eff}}feff​ of this system can be found using the mirror formula and lens formula combined, but given the scenario, we focus on this system.

2. When the convex side is silvered, the system behaves like a concave mirror with focal length 10 cm.

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 refraction takes place at both the air-glass interface and glass-air interface of a rectangular glass slab. When the light ray incident on the air-glass interface (DC) obliquely, it bends towards the normal.

Hypermetropia is corrected using converging lens.  In this defect the person is able to see far objects clearly but difficulty with near vision. The image is focused behind the retina rather than upon it. This occurs when eyeball is too short or the power of lens is too weak. By using a convex lens the defect is corrected.

We have, (1/v)−(1/−90)=1/f⇒(1.5−1)(1/30)=1/60
∴ v=180 cm
∴|m|=∣v/u∣=180/90=2.0

The only difference in covering any part of a lens is that the intensity of the image will reduce. Since the focal length is a function of curvature of the surfaces of the lens and covering a part does not change it, the image will be formed as usual.

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