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