Ray Optics and Optical Test

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Ray Optics and Optical Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

As shown in figure, the liquids $$L_1,L_2$$ and $$L_3$$ have refractive indices. $$1.55,\ 1.50$$ and $$1.20$$ respectively. Therefore, the arrangement corresponds to

A
biconvex lens

B
biconcave lens

C
concavo-convex lens

D
convexo-concave lens
Question 2
1 / -0

A ray of light travels from air to liquid by making an angle of incidence $$24^o$$ and angle of refraction is $$18^o$$. Find the R.I. of liquid. Determine the wavelength of light in air and in liquid the frequency of light is $$5.4\times 10^{15}$$ Hz. c$$=3\times 10^8$$ m/s.

A
$$2.316, 5656$$A; $$422.13\overset{o}{A}$$.

B
$$2.316, 5556$$A; $$422.13\overset{o}{A}$$.

C
$$1.316, 5656$$A; $$422.13\overset{o}{A}$$.

D
$$1.316, 5556$$A; $$422.13\overset{o}{A}$$.

Solution

Given angle of incidence $${\theta}_{i}$$=$${24}^{0}$$ and angle of refraction $${\theta}_{r}$$=$${18}^{0}$$ .
Also frequency of light in liquid =5.4$$\times {10}^{15}$$ and velocity of light in air =3 $$\times {10}^{8}$$.
From Snell's law,
$$\dfrac { \sin{ 24 }^{ 0 } }{ \sin18^{ 0 } } =\dfrac { { \mu }_{ l } }{ { \mu }_{ a } } $$
Thus refractive index of liquid $$=\dfrac { 0.4067 }{ 0.30901 } =1.3161$$
Also $$\dfrac { { \mu }_{ l } }{ { \mu }_{ a } } =\dfrac { { v }_{ a } }{ { v }_{ l } } $$----(A)
We know frequency f$$=\dfrac { v }{ \lambda } $$
So equation A can be written as
$$1.316=\dfrac { 3\times { 10 }^{ 8 } }{ 5.4\times { 10 }^{ 15 }\times { \lambda }_{ L } } \\ { \lambda }_{ L }=\dfrac { 3\times { 10 }^{ 8 } }{ 5.4\times { 10 }^{ 15 }\times 1.316 } =422.53\times { 10 }^{ -10 }$$
Question 3
1 / -0

a diverging lens of focal length - $$10$$ cm is moving towards right with a velocity $$5$$ m/s. An object, placed on Principal axis is moving towards left with a velocity $$3$$ m/s. The velocity of the image at the instant when the lateral magnification produced is $$1/2$$ is: (All velocities are with respect to ground)

A
$$3$$ m/s towards right

B
$$3$$ m/s towards left

C
$$7$$ m/s towards right

D
$$7$$ m/s towards left

Solution

$$\begin{array}{l} f=-10\, cm \\ M=\frac { 1 }{ 2 } =\left( { \frac { v }{ u } } \right) \\ { v_{ IM } }=\left( { \frac { { { v^{ 2 } } } }{ { { u^{ 2 } } } } } \right) { v_{ OM } } \\ \Rightarrow { v_{ I } }-{ v_{ M } }=\frac { 1 }{ 4 } \left( { { v_{ O } }-{ v_{ M } } } \right) \\ { v_{ I } }-5=\frac { 1 }{ 4 } \left[ { -3-5 } \right] \\ { v_{ I } }-5=-2 \\ { v_{ I } }=3\, m/s\, \, \, \, towards\, right \end{array}$$
Hence,
option $$(A)$$ is correct answer.
Question 4
1 / -0

A parallel beam of width $$'a'$$, is incident on the surface of glass slab $$(\mu=3/2)$$, at an angle $$'i'$$ and the angle of refraction in glass is $$'r'$$. The width of the refracted parallel beam will be ?

A
equal to $$a$$

B
less than $$a$$

C
more than $$a$$

D
exactly $$2a$$

Solution
Question 5
1 / -0

A ray of light of wave length $$4500$$ A.U. in water is refreacted in air and passes just grazing the water surface. Find the angle of incience of ray in water if wavelength of light in air is $$6000$$ A.U.

A
$$i = 58^o 36'$$

B
$$i = 48^o 36'$$

C
$$i = 68^o 36'$$

D
$$i = 88^o 36'$$
Question 6
1 / -0

A rectangular glass slab $$ABCD$$ of refractive index $$n_{1}$$ is immersed in water of refractive index $$n_{2}(n_{1} < n_{2})$$. A ray of light is incident at the surface $$AB$$ of the slab as shown. The maximum value of the angle of incidence $$\alpha_{max}$$ such that the ray comes out from the other surface $$CD$$ is given by

A
$$\sin^{-1}\left [\dfrac {n_{1}}{n_{2}}\cos \left (\sin^{-1} \dfrac {n_{2}}{n_{1}} \right )\right ]$$

B
$$\sin^{-1}\left [n_{1} \cos \left (\sin^{-1} \dfrac {1}{n_{2}}\right )\right ]$$

C
$$\sin^{-1}\left (\dfrac {n_{1}}{n_{2}}\right )$$

D
$$\sin^{-1}\left (\dfrac {n_{2}}{n_{1}}\right )$$

Solution

$$\dfrac { { n }_{ 1 } }{ { n }_{ 2 } } =\dfrac { { sin\sigma }_{ max } }{ { sin }_{ r1 } } ={ \sigma }_{ max }={ sin }^{ -1 }\left[ \dfrac { { n }_{ 1 } }{ { n }_{ 2 } } sin{ r }_{ 1 } \right] $$

$${ r }_{ 1 }+{ r }_{ 2 }={ 90 }^{ \circ }\Rightarrow { r }_{ 1 }=90-{ r }_{ 2 }=90-C$$

$${ r }_{ 1 }=90-{ sin }^{ -1 }\left( \dfrac { 1 }{ { 2\mu }_{ 1 } } \right) \Rightarrow { r }_{ 1 }=90-{ r }_{ 2 }=90-C$$

$${ \sigma }_{ max }={ sin }^{ -1 }\left[ \dfrac { { n }_{ 1 } }{ { n }_{ 2 } } sin\left\{ 90-{ sin }^{ -1 }\dfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right\} \right] $$

$$={ sin }^{ -1 }\left[ \dfrac { { n }_{ 1 } }{ { n }_{ 2 } } cos\left( { sin }^{ -1 }\dfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \right] $$
Question 7
1 / -0

A cubical transparent slab is used as a paper weight.What should be the minimum refractive index of the material of the slab so that letters below it are not visible from any of its vertical faces.

A
$$\sqrt 2 $$

B
$$\sqrt 3 $$

C
$$\sqrt 5 $$

D
$$\sqrt 5 $$$$\sqrt 3 $$

Solution
Question 8
1 / -0

The graph between u and v for a convex mirror is

A

B

C

D
Question 9
1 / -0

A ray of light travels from water to glass as shown below. The refractive index of water is $$1.3$$ and the refractive index of glass is $$1.5$$. What is the angle of refraction ?

A
$$30.7^{o}$$

B
$$35.3^{o}$$

C
$$41.7^{o}$$

D
$$48.6^{o}$$

Solution
Question 10
1 / -0

A transparent solid cylindrical rod has a refractive index of $$\frac{2}{{\sqrt 3 }}$$. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in figure:
The incident angle $$\theta$$ for which the light ray grazes along the wall of the rod is:

A
$${\sin^{-1}}(\frac{1}{2})$$

B
$${\sin^{-1}}(\frac{\sqrt3}{2})$$

C
$${\sin^{-1}}(\frac{2}{\sqrt3})$$

D
$${\sin^{-1}}(\frac{1}{\sqrt3})$$