Ray Optics and Optical Test

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Ray Optics and Optical Test - 64

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Weekly Quiz Competition

Question 1
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The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called ...............

A
Centre of curvature

B
Radius of curvature

C
Focus

D
Pole

Solution

The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror. Here, the distance $$R$$ is the radius of curvature in the figure.

Answer-(B).
Question 2
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A ray of light passes through four transparent media with refractive indices$$\mu_1, \mu_2, \mu_3$$ and $$\mu_4$$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have:

A
$$\mu_1=\mu_2$$

B
$$\mu_2=\mu_3$$

C
$$\mu_3=\mu_4$$

D
$$\mu_4=\mu_1$$

Solution

We know that $$\mu \sin\theta=constant$$
where $$\mu=$$ refractive index of medium
and $$\theta=$$ angle between direction of propagation and normal

Here, for medium -1 and medium-4:-
rays are parallel , and hence $$\theta$$ will be same.

$$\implies\mu_1=\mu_4$$

Answer-(D)
Question 3
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A glass slab of thickness 18 cm and refractive index $$\dfrac{3}{2}$$ is placed on a printed matter. The normal shift of the printed matter is

A
3 cm

B
6 cm

C
9 cm

D
12 cm

Solution

We know that, Refractive index is $$\mu = \displaystyle \frac{\textrm{Real Depth}}{\textrm{Apparent Depth}}$$

Given the Real depth $$d_\textrm{real} = 18\textrm{ cm}$$
Refractive index $$\mu = 1.5$$

Thus, Apparent depth $$d_\textrm{apparent} = \displaystyle \frac{d_\textrm{real}}{\mu} = \frac{18}{1.5} = 12 \textrm{ cm}$$

Thus Normal Shift = Real Depth - Apparent Depth$$ = 18 - 12 = 6\textrm{ cm}$$
Question 4
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The process of re-emission of absorbed light in all directions with different intensities by the atom or molecule is called ____________.

A
Scattering of light

B
Dispersion of light

C
Reflection of light

D
Refraction of light

Solution

The process of absorbed light in all possible directions with different intensities and frequencies by an atom or a molecule is called the scattering of light. When a beam of light strikes such fine particles, the path of the beam becomes visible. The light reaches us, after being reflected diffusely by these particles.
Question 5
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A rectangular block is composed of three different glass prisms (with refractive indices $$\mu_1, \mu_2$$ and $$\mu_3$$) as shown in the figure below. A ray of light incident to the left face emerges normal to the right face. Then the refractive indices are related as:

A
$$\mu_1^2 + \mu_2^2 = 2 \mu_3^2$$

B
$$\mu_1^2 + \mu_2^2 = \mu_3^2$$

C
$$\mu_1^2 + \mu_3^2 = 2 \mu_2^2$$

D
$$\mu_2^2 + \mu_3^2 = 2 \mu_1^2$$

Solution

From the figure,
$$i_1 = r_2 = 45^o$$

In $$\Delta AEF,$$,
$$\angle A + \angle E + \angle F = 180^o$$
$$\implies \angle A = 90^o$$

In quadrilateral $$ABCD,$$
$$\angle A + \angle B + \angle C + \angle D = 180^o$$
$$\implies \angle C = 90^o$$

in $$\Delta BCD,$$
$$\angle B + \angle C + \angle D = 180^o$$
$$r_1 + i_2 = 90^o$$
$$\sin r_1 = \cos i_2$$
$$\sin^2 r_1 = \cos^2 i_2$$
$$\sin^2 r_1 = 1-\sin^2 i_2$$
Using snell's law,
$$\cfrac{\sin^2 i_1}{(\mu_2/\mu_1)^2} = 1 - (\mu_3/\mu_2)^2 \sin^2 r_2$$
$$\cfrac{\mu_1^2}{2\mu_2^2} + \cfrac{\mu_3^2}{2\mu_2^2}=1$$
$$\mu_1^2 + \mu_3^2 = 2\mu_2^2$$
Question 6
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A vessel is filled with oil as shown in the diagram. A ray of light from point $$O$$ at the bottom of vessel is incident on the oil - air interface at point $$P$$ and grazes the surface along $$PQ$$. The refractive index of the oil is close to-

A
$$1.41$$

B
$$1.50$$

C
$$1.630$$

D
$$1.73$$

Solution

$$\sin { { i }_{ c } } =\cfrac { 12 }{ \sqrt { { (17) }^{ 2 }+{ (12) }^{ 2 } } } =\cfrac { 12 }{ 20.80 } $$
$$\quad \mu =\cfrac { 1 }{ \sin { { i }_{ c } } } =\cfrac { 20.80 }{ 12 } =1.73\quad \quad $$
Question 7
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Two glass prisms $$P_1$$ and $$P_2$$ are to be combined together to produce dispersion without deviation. The angles of the prisms $$P_1$$ and $$P_2$$ are selected as $$4^o$$ and $$3^o$$ respectively. If the refractive index of prism $$P_1$$ is 1.54, then that of $$P_2$$ will be

A
1.48

B
1.58

C
1.62

D
1.72

Solution

Prism angle of first prism $$A_1 =4^o $$
Its refractive index $$\mu_1=1.54$$
Prism angle of second prism prism $$A_2 =3^o $$
Let refractive index of second prism be $$\mu_2$$.
For dispersion without deviation, net deviation of the ray must be zero i.e. $$\delta_1 - \delta_2 =0$$
$$\therefore$$ $$(\mu_1 -1)A_1 -(\mu_2 -1)A_2 =0$$
OR $$(1.54 -1)(4) -(\mu_2 -1)3 =0$$
OR $$\mu_2 -1 =0.72$$ $$\implies \mu_2 = 1.72$$
Question 8
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Figure shows an irregular block of material of refractive index $$\sqrt { 2 }$$. A ray of light strikes the face $$AB$$ as shown in figure. After refraction, it is incident on a spherical surface $$CD$$ of radius of curvature $$0.4 m$$ and enters a medium of refractive index $$1.514$$ to meet $$PQ$$ at $$E$$. Find the distance $$OE$$ upto two places of decimal :

A
$$7 m$$

B
$$.29 m$$

C
$$6.06 m$$

D
$$8.55 m$$

Solution

From Snell's law,
$${ \mu }_{ 1 }\sin { i } ={ \mu }_{ 2 }\sin { r } $$
$$\Rightarrow \sin { r } =\dfrac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } \sin { i } =\dfrac { 1 }{ \sqrt { 2 } } \sin { { 45 }^{ o } } $$
$$\Rightarrow \dfrac { 1 }{ 2 } =\sin { r } \Rightarrow r={ 30 }^{ o }$$
This means that the ray becomes parallel to side $$AD$$ inside the slab.
This implies for the second face $$CD$$, $$u=\infty $$
Given $$R=0.4 m$$
$$\dfrac { { \mu }_{ 3 } }{ v } -\dfrac { { \mu }_{ 2 } }{ v } =\dfrac { { \mu }_{ 3 }-{ \mu }_{ 2 } }{ R } $$
$$\Rightarrow \dfrac { 1.514 }{ v } -\dfrac { \sqrt { 2 } }{ \infty } =\dfrac { 1.514-\sqrt { 2 } }{ R } $$
$$v=\dfrac { 1.514\times 0.4 }{ 1.514-1.414 } =\dfrac { 1.514\times 0.4 }{ 0.1 } $$
$$v=6.056=6.06 m$$ (upto $$2$$ decimal places).
Question 9
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A ray of light passes through four transparent media with refractive index $${n}_{1}, {n}_{2}, {n}_{3}$$ and $${n}_{4}$$ as shown. The surfaces of all media are parallel.
If the emergent ray $$DE$$ is parallel to the incident ray $$AB$$, then:

A
$${ n }_{ 1 }={ n }_{ 4 }$$

B
$${ n }_{ 2 }={ n }_{ 4 }$$

C
$${ n }_{ 3 }={ n }_{ 4 }$$

D
$${ n }_{ 1 }=\dfrac { { n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 } }{ 3 } $$

Solution

As the emergent ray $$DE$$ is parallel to incident ray $$AB$$, thus angle of incidence $$i$$ is equal to angle of refraction $$r$$.
$$\implies$$ $$i=r$$ ...........(1)
Using Snell's law of refraction,
$$n_1\sin i = n_4 \sin r$$
$$\therefore$$ $$n_1 \sin i = n_4 \sin i$$ (using 1)
$$\implies n_1 = n_4$$
Question 10
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A man is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is $$ 1.39 \times 10^9 m$$ and its mean distance from the earth is $$1.5 \times 10^{11} \ m$$, the diameter of the sun's image on the paper is

A
$$3.1 \times 10^{-4} m$$

B
$$6.5 \times 10^{-5} m$$

C
$$6.5 \times 10^{-4} m$$

D
$$9.2 \times 10^{-4} m$$

Solution

Magnification of convex lens, $$m \, = \, \dfrac{-f}{f \, - \, u}$$
Here, $$f \, = \, 10 \, cm \, = \, 10 \, \times \, 10^{-2} \, m, \, \, u \, = \, 1.5 \, \times \, 10^{11} \, m$$
$$\therefore \, m \, = \, \dfrac{-10 \, \times \, 10^{-2}}{10 \, \times \, 10^{-2} \, - \, 1.5 \, \times \, 10^{11}} \, = \, 6.67 \, \times \, 10^{-13}$$
$$\therefore$$ Diameter of the image,
$$d \, = \, m \, \times \, 1.39 \, \times \, 10^9 \, = \, 6.67 \, \times \, 10^{-13} \, \times \, 1.39 \, \times \, 10^9 \, = \, 9.27 \, \times \, 10^{-4} \, m$$
Hence, the correct option is $$(D)$$