Ray Optics and Optical Test

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Ray Optics and Optical Test - 63

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Weekly Quiz Competition

Question 1
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Three identical isosceles right-angled prism (ABC, BCD and CDE) are placed as shown in the figure $\displaystyle \angle ABC,\angle BCD\, and\, \angle CDE$ are $\displaystyle 90^{\circ}$ in the given prism respectively. A ray is incident along line AB Refraction takes placed at BC and CD and then it emerges along the line DE $\displaystyle \mu _{1},\mu _{2}$ and $\displaystyle \mu _{3}$ are the refractive index of prism ABC, BCD and CDE respectively. The relation between $\displaystyle \mu _{1},\mu _{2}\,and\,\mu _{3}$ is

A
$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-\mu _{2}^{3}+2=0$

B
$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-\mu _{3}^{2}+2=0$

C
$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-2\mu _{3}^{2}+2=0$

D
$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-5\mu _{3}^{2}+2=0$

Solution

$c=$ critical angle with respect $\mu_{1}$ (Prism $\left.A B C\right)$
$\mu_{1} \sin i_{1}=M_{2} \sin \gamma_{c}$
at surface $A B$
(1) $\sin (90)=M, \sin C$
$\frac{1}{M_{1}}=\sin c$
$\mu_{1} \sin (90-c)=\mu_{2} \sin \gamma_{1}$
$\mu_{1} \cos c=\mu_{2} \sin \gamma_{1}$
$\mu_{1} \frac{\sqrt{\mu_{1}^{2}+1}}{\mu_{1}}=\mu_{2} \sin \gamma_{1}$
$\mu_{1}^{2}+1=m_{2}^{2} \sin ^{2} \gamma_{1}-$ (3)
at surface $x$
$\begin{array}{c}\mu_{2} \sin \left(q 0\gamma_{2}\right)=\mu_{3}\sin\gamma_{2}\\\mu_{2}\cos\gamma_{1}=\mu_{3} \sin \gamma_{2}\end{array}-(1)$
at surface $D \in$
$\begin{aligned}\mu_{3}\sin\left(90\gamma_{2}\right) &=1 \\ \cos \gamma_{2} &=\frac{1}{\mu_{3}}\end{aligned}-(2)$
$\frac{\mu_{2}^{2} \cos ^{2} \gamma_{1}}{\mu_{3}^{2}}=\sin ^{2} \gamma_{2} \quad$ Squaring equation
$+\frac{1}{\mu_{3}^{2}}=\cos ^{2} \gamma_{2}$ sequaring equation
$\mu_{2}^{2} \cos ^{2} \gamma_{1}+1=\mu_{3}^{2}$
Substitute (3) in this
$\mu_{2}^{2}\left[\frac{\mu_{1}^{2}+1}{\mu_{2}^{2}}\right]+1=\mu_{3}^{2}$
$\mu_{1}^{2}-\mu_{3}^{2}+2=0$
Question 2
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Three right angled prisms of refractive indices $n_1, n_2$ and $n_3$ are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

A
$n_1 = n_2 = n_3$

B
$n_1 = n_2 \neq n_3$

C
$1 + n_1 = n_2 + n_3$

D
$1 + n_2^2 = n_1^2 + n_3^2$

Solution

Applying Snell's Law at each surface,

AT B, $sin i=n_1sin r_1 \implies sin^2 i=n_1^2 sin^2r_1$

At C, $n_1sin(90-r_1)=n_2 sin r_2 \implies n_1^2cos^2 r_1=n_2^2sin^2 r_2$

At D, $n_2 sin(90-r_2)=n_3 sinr_3 \implies n_2^2 cos^2r_2=n_3^2 sin^2r_3$

At E, $n_3 sin(90-r_3)=sin(90-i) \implies n_3^2cos^2r_3=cos^2i$

Adding all above equations gives,

$1+n_2^2=n_1^2+n_3^2$
Question 3
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Directions For Questions

The figure shows a prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass it is equivalent to two $30^o-60^o-90^o$ prisms and one $45^o-45^o-90^o$ Prism.
The angle $\theta_1$ is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. Refractive index for the prism, $\\2 sin \theta_1$ .

...view full instructions

The total deviation of the incident ray when it emerges out of the prism is.

A
$30^o$

B
$60^o$

C
$90^o$

D
$45^o$

Solution

In figure (a), the incident ray $P$ hits the point Q, R and S and then emerges as $T$ . The overall scenario is seen in figure (b) which represents the deviation as $\delta$ .
$\therefore$ Deviation $\delta = 90- \theta_2 + \theta_1$ .............(1)
In the prism $DAC$ , prism angle $= 30^o$
Using prism angle formula $A = r_2+ r'$ , $\therefore$ $30^0 = r_1 + 0^o \implies r_2 = 30^o$
Given: Refractive index of the prism $n = 2sin\theta_1$

Using Snell's law at point $S$ , $1 \times sin\theta_2 = n \times sinr_2$
$\therefore$ $sin\theta_2 = 2sin\theta_1 \times sin30$
$sin\theta_2 = sin\theta_1$    $\implies$ $\theta_1 = \theta_2$
$\therefore$ From equation (1), deviation of the ray $\delta = 90^o$
Question 4
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Directions For Questions

The figure shows a prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass it is equivalent to two $30^o-60^o-90^o$ prisms and one $45^o-45^o-90^o$ Prism.
The angle $\theta_1$ is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. Refractive index for the prism, $\\2 sin \theta_1$ .

...view full instructions

The ratio of $\frac {\theta _1}{\theta _2}$ is.

A
1

B
$\frac {1}{2}$

C
$\sqrt 2$

D
$\frac {1}{\sqrt 2}$

Solution

When ray PQ enters in prism, by Snell's law:
$\dfrac{\sin\theta_{1}}{\sin r_{1}}=\mu=2\sin\theta_{1}$ .......eq1
$\sin r_{1}=1/2=\sin30^{o}$
$r_{1}(\angle RQN)=30^{o}$
$\angle RQN'=90-30=60^{o}$

Now BMC is $45^{o}=45^{o}-90^{o}$ prism ( $\angle M=90$ )
Hence, $\angle B=30+45=75^{o}$

In $\triangle BQR$
$\angle BRQ=180-(75+60)=45^{o}$

Hence, $\angle QRE=90-45=45^{o}$
In quadrilateral QRSA ,
$\angle Q=90+30=120^{o}$
$\angle A=30+60=90^{o}$
$\angle R=45+45=90^{o}$
Hence, $\angle ASR=360-(120+90+90)=60^{o}$
It gives $\angle RSK=90-60=30^{o} =r_{2}$

When ray ST emerges from prism, by Snell's law
$\dfrac{\sin\theta_{2}}{\sin r_{2}}=\dfrac{1}{\mu}$ ................eq2

Dividing eq1 by eq2 ,
$\dfrac{\theta_{1}}{\theta _{2}}=1$ (as $r_{1}=r_{2}=30^{o}$ )
Question 5
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There is an equiconvex lens of focal length of $20 cm$ . If the lens is cut into tow equal parts perpendicular to the optical axis, the focal length of each part will be

A
20 cm

B
10 cm

C
40 cm

D
15 cm

Solution

The focal length of equiconvex lens is 20 cm
$\frac { 1 }{ f } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \\ for\quad equiconvex\quad lens\quad { R }_{ 1 }=-{ R }_{ 2 }=R\\ thus,\\ \frac { 1 }{ 20 } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } } +\frac { 1 }{ { R } } \right) \\ \Rightarrow R=40\left( \mu -1 \right)$
when lens is cut into two equal parts perpendicular to optical axis
then,
${ R }_{ 1 }=\infty ,\quad { R }_{ 2 }=R$
thus
$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right)$
$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { R } } \right)$
by putting value of R we get,
$\frac { 1 }{ f' } =\left( \mu -1 \right) \left( \frac { 1 }{ { 40\left( \mu -1 \right) } } \right) \\ \Rightarrow f'=40cm$
new focal length of each part is 40 cm , becomes double

Option C is correct.
Question 6
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A lens is made of flint glass (refractive index = 1.5). When the lens is immersed in a liquid of refractive index 1.25, the focal length :

A
Increases by a factor of 1.25

B
Increases by a factor of 2.5

C
Increases by a factor of 1.2

D
Remains same

Solution

From len's makers formula, $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$

$\dfrac{1}{f_a}=(1.5-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$

$\dfrac{1}{f_l}=(\mu_g-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$

$^lu_g = \dfrac{\mu_g}{\mu_l} = \dfrac{1.5}{1.25}=\dfrac{6}{5}$

$\dfrac{1}{f_l}=\left(\dfrac{6}{5}-1\right)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)=\dfrac{1}{5}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$

$\dfrac{1/f_a}{1/f_l}=\dfrac{0.5}{1/5}$

$\Rightarrow \dfrac{f_l}{f_a}=0.5\times 5=2.5$

$f_l=2.5\times f_a$
Question 7
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The diameter of the reflecting surface of spherical mirror is called ..............

A
Focus

B
Aperture

C
Centre of curvature

D
None of these

Solution

The diameter of the reflecting surface of the spherical mirror is called it's $aperture$ .
Question 8
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A thin prism $P_1$ with angle $6^o$ and made from glass of refractive index 1.54 is combined with another thin prism $P_2$ of refractive index 1.72 to produce dispersion without deviation. The angle of prism $P_2$ will be.

A
$4^o 30'$

B
$8.5^o$

C
$6.5^o$

D
None of these

Solution

We know that for small angle of prism, deviation , $\delta=A(\mu-1)$
where $A$ =angle of prism

Here, net deviaton $A_1(\mu_1-1)-A_2(\mu_2-1)=0$

$\implies A_2=\dfrac{A_1(\mu_1-1)}{\mu_2-1}$

$\implies A_2=\dfrac{6\times 0.54}{0.72}=4.5^{o}$

$\implies A_2=4^{o}30'$

Answer-(A)
Question 9
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Height of the object and height of the image are measured from the:

A
Optical centre

B
Principal axis

C
Principal focus

D
Aperture

Solution

According to convention height of object and image are measured from principal axis.

Answer-(B)
Question 10
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A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45 $^o$ and that of refraction is 30 $^o$ . Find the refractive index of the medium.

A
2

B
$\dfrac{1}{\sqrt 2}$

C
$2 \sqrt 2$

D
$\sqrt 2$

Solution

Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by $\displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r}$

The initial medium is air. Thus $\mu_1 = 1$
Given incidence angle $i = 45^o$ and refraction angle is $30^o$

By Snell's law, $\mu_2 = \mu_1\times \displaystyle \frac{\sin i}{\sin r} = 1\times \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Test - 63

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Ray Optics and Optical Test - 63

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Question 1

μ22+μ12−μ23+2=0\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-\mu _{2}^{3}+2=0μ22​+μ12​−μ23​+2=0

μ22−μ12−μ32+2=0\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-\mu _{3}^{2}+2=0μ22​−μ12​−μ32​+2=0

μ22+μ12−2μ32+2=0\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-2\mu _{3}^{2}+2=0μ22​+μ12​−2μ32​+2=0

μ22−μ12−5μ32+2=0\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-5\mu _{3}^{2}+2=0μ22​−μ12​−5μ32​+2=0

Solution

Question 2

n1=n2=n3n_1 = n_2 = n_3n1​=n2​=n3​

n1=n2≠n3n_1 = n_2 \neq n_3n1​=n2​=n3​

1+n1=n2+n31 + n_1 = n_2 + n_31+n1​=n2​+n3​

1+n22=n12+n321 + n_2^2 = n_1^2 + n_3^21+n22​=n12​+n32​

Solution

Question 3

30o30^o30o

60o60^o60o

90o90^o90o

45o45^o45o

Solution

Question 4

1

12\frac {1}{2}21​

2\sqrt 22​

12\frac {1}{\sqrt 2}2​1​

Solution

Question 5

20 cm

10 cm

40 cm

15 cm

Solution

Question 6

Increases by a factor of 1.25

Increases by a factor of 2.5

Increases by a factor of 1.2

Remains same

Solution

Question 7

Focus

Aperture

Centre of curvature

None of these

Solution

Question 8

4o30′4^o 30'4o30′

8.5o8.5^o8.5o

6.5o6.5^o6.5o

None of these

Solution

Question 9

Optical centre

Principal axis

Principal focus

Aperture

Solution

Question 10

2

12\dfrac{1}{\sqrt 2}2​1​

222 \sqrt 222​

2\sqrt 22​

Solution

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$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-\mu _{2}^{3}+2=0$

$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-\mu _{3}^{2}+2=0$

$\displaystyle \mu _{2}^{2}+\mu _{1}^{2}-2\mu _{3}^{2}+2=0$

$\displaystyle \mu _{2}^{2}-\mu _{1}^{2}-5\mu _{3}^{2}+2=0$

$n_1 = n_2 = n_3$

$n_1 = n_2 \neq n_3$

$1 + n_1 = n_2 + n_3$

$1 + n_2^2 = n_1^2 + n_3^2$

$30^o$

$60^o$

$90^o$

$45^o$

$\frac {1}{2}$

$\sqrt 2$

$\frac {1}{\sqrt 2}$

$4^o 30'$

$8.5^o$

$6.5^o$

$\dfrac{1}{\sqrt 2}$

$2 \sqrt 2$

$\sqrt 2$