Ray Optics and Optical Test

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Ray Optics and Optical Test - 50

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Question 1
1 / -0

A liquid of refractive index 1.6 is introduced between two identical plano-convex lens in two ways P and Q as shown. If the lens material has refractive index 1.5, the combination is

A
converging in both P and Q

B
diverging in both P and Q

C
converging in Q only

D
converging in P only

Solution
Question 2
1 / -0

Maximum lateral displacement of a ray of light incident $$( i =90)$$ on a slab of thickness $$t$$ is:

A
$$\dfrac{t}{2}$$

B
$$\dfrac{t}{3}$$

C
$$\dfrac{t}{4}$$

D
$$t$$

Solution

We know that,

For $$i={{90}^{0}}$$

Lateral shift is maximum

Lateral shift for a slab of t thickness

$$ L.S=t\times \dfrac{\sin \left( i-r \right)}{\cos r} $$

$$ L.S=t\times \dfrac{\sin \left( 90-r \right)}{\cos r} $$

$$ L.S=t\times \dfrac{\cos r}{\cos r} $$

$$ L.S=t $$

Hence, the lateral shift is equal to thickness $$t$$
Question 3
1 / -0

A light ray is incident normally on the surface $$AB$$ of a prism of refracting angle $$60^\circ$$. If the light ray does not emerge from $$AC$$, then find the refractive index of the prism.

A
$$\mu \geqslant \cfrac{\sqrt{3}}{2}$$

B
$$\mu \leqslant \cfrac{\sqrt{3}}{2}$$

C
$$\mu \geqslant \cfrac{2}{\sqrt{3}}$$

D
$$\mu \leqslant \cfrac{2}{\sqrt{3}}$$

Solution

Here, for no emergence through AC, TIR happens at AC.
$$\therefore \mu \sin i\ge 1\times \sin 90°$$

or $$\mu \ge \dfrac{1}{\sin i}$$

or $$\mu \ge \dfrac{1}{\sin 60°}$$

or $$\mu \ge \dfrac{2}{\sqrt{3}}$$
Question 4
1 / -0

A thin concavo-concave lens is surrounded by two different liquids A and B as shown in figure. The system is supported by a plane mirror at the bottom. Refractive index of A, lens and B are 9/5 , 7/5 and 8/7 respectively.The radius of curvature of the surfaces of the lens are same and equal to 10 cm. Where should an object be placed infront of this system so that final image is formed on the object itself.

A
60 cm

B
75 cm

C
50 cm

D
None
Question 5
1 / -0

A beaker contains Water and Benzene of refractive indices $$\dfrac{4}{3}$$ and $$\dfrac{3}{2}$$ respectively one above the other. If depth of each is $$12 \ cm$$, apparent depth of the bottom of the beaker is

A
$$17 \ cm$$

B
$$16 \ cm$$

C
$$18 \ cm$$

D
$$19 \ cm$$

Solution

Apparent depth= $$\cfrac {12}{\mu _1}+\cfrac {12}{\mu_2}$$
$$=\cfrac {12 \times 3}{4}+\cfrac {12\times 2}{3}$$
$$=9+8$$
$$=17 cm$$
Question 6
1 / -0

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describe best the image formed of an object of height 2 cm placed 30 cm from the lens?

A
Real, inverted, height = 1 cm

B
Virtual, upright, height = 1 cm

C
Virtual, upright, height = 0.5 cm

D
Real, inverted, height =4 cm

Solution
Question 7
1 / -0

A Plano convex lens is made of glass of refractive index 1.5.The of curvature of its convex surface is R. Its focal length:

A
R/2

B
R

C
2R

D
1.5 R

Solution

Given,

Radius of curvature, $$R$$
Refractive index = $$ 1.5 $$

Focal length of lens, $$f=\dfrac{R}{(\mu-1)}=\dfrac{R}{(1.5-1)}=2R$$
Question 8
1 / -0

The refractive index of flint glass for blue F line is $$1.6333$$ and for red C line is $$1.6161$$. If the refractive index for yellow D line is $$1.622$$, then the dispersive power of glass is

A
$$0.0276$$

B
$$0.276$$

C
$$2.76$$

D
$$0.106$$

Solution

We have the equation of dispersive power,

$$\omega=\dfrac{\mu_F - \mu_C}{(\mu_D-1)}$$

where,

$$\mu_F$$ is the refractive index of blue $$F$$ line

$$\mu_C$$ is the refractive index of blue $$C$$ line

$$\mu_D$$ is the refractive index of blue $$D$$ line

Then,

$$\omega=\dfrac{1.6333-1.6161}{1.622-1}=0.0276$$
Question 9
1 / -0

A lens forms a vertical image 14 cm away from it when an object is placed 10 cm away from it. The lens is a....... less of focal length.......

A
concave , 6.67 cm

B
concave , 2.86 cm

C
convex , 2.86 cm

D
none

Solution
Question 10
1 / -0

Time taken by the sunlight to pass through a window of thickness 4mm whose refractive index is 1.5 is

A
$$3 \times 10^{-8}s$$

B
$$2 \times 10^{-8}s$$

C
$$3 \times 10^{-11}s$$

D
$$2 \times 10^{-11}s$$

Solution

Given,
Refractive index of medium = 1.5
Thickness of window $$=4 \ mm=4\times 10^{-3} \ m$$
Velocity of light in the window
$$=\dfrac {3\times 10^8}{1.5}ms^{-1}=2\times 10^8 ms^{-1}$$

Hence $$t=\dfrac {4\times 10^{-3}}{2\times 10^8}s=2\times 10^{-11}s$$