Question 371: Ray Optics

Question

For an isosceles prism of angle $A$ and refractive index $\mu $, it is found that the angle of minimum deviation ${\delta _m} = A.$
Which of the following options is/are correct?

Accepted Answer

We discuss the options as follows:
$\bullet$ For option (A) : For the angle of incidence i
1
= A, we have minimum deviation and the ray inside the prism is parallel to the base of the prism.
Hence, option (A) is correct.
$\bullet$ For option (C) : The angle of deviation is given as
$\delta$ = i
1
+ i
2
$-$ A
At minimum deviation, i
1
= i
2
and r
1
= r
2
. Therefore,
i
1
= A ....... (1)
Also, we know that
r
1
+ r
2
= A
As r
1
= r
2
, we get
2r
1
= A $\Rightarrow$ r
1
= ${A \over 2}$
From Eq. (1), we get
${r_1} = {{{i_1}} \over 2}$
Hence, option (C) is correct.
According to Snell's law, we have
${{\sin {i_2}} \over {\sin {r_2}}} = \mu $
For emergent ray to be tangential to the surface, we have
i
2
= 90$^\circ$
$\Rightarrow$ sin90$^\circ$ = $\mu$sinr
2
$\Rightarrow$ $\mu$sinr
2
= 1
sinr
2
= ${1 \over \mu }$ $\Rightarrow$ r
2
= sin
$-$1
$\left( {{1 \over \mu }} \right)$ ....... (2)
From the relation r
1
+ r
2
= A, we get
r
1
= A $-$ r
2
$\Rightarrow$ sinr
1
= sin(A $-$ r
2
)
Using sin(a $-$ b) = sinacosb $-$ sinbcosa, we have
$\Rightarrow$ sinr
1
= sinAcosr
2
$-$ sinr
2
cosA ......... (3)
Using Eq. (2), we have
sinr
2
= ${{1 \over \mu }}$
Squaring it, we get
sin
2
r
2
= ${{1 \over {{\mu ^2}}}}$
Using the relation sin
2
x + cos
2
x = 1, we get
sin
2
x = 1 $-$ cos
2
x
$\Rightarrow$ 1 $-$ cos
2
r
2
= ${{1 \over {{\mu ^2}}}}$
$\Rightarrow$ cos
2
r
2
= 1 $-$ ${{1 \over {{\mu ^2}}}}$
That is,
cosr
2
= $\sqrt {1 - {1 \over {{\mu ^2}}}} $
Substituting the values of sinr
2
and cosr
2
in Eq. (3), we get
sinr
1
= sin A$\sqrt {1 - {1 \over {{\mu ^2}}}} $ $-$ ${{1 \over \mu }}$cos A
According to Snell's law, we have
${{\sin {i_1}} \over {\sin {r_1}}} = \mu $
That is,
$\sin {i_1} = \mu \sin {r_1} = \mu \left( {\sin A\sqrt {1 - {1 \over {{\mu ^2}}}}  - {1 \over \mu }\cos A} \right)$
$\sin {i_1} = \mu \left( {\sin A{{\sqrt {{\mu ^2} - 1} } \over \mu } - {{\cos A} \over \mu }} \right)$
$\sin {i_1} = \sin A\sqrt {{\mu ^2} - 1}  - \cos A$ ....... (4)
Now, at minimum deviation, we have A = i
1
and r
1
= ${A \over 2}$. Thus, from Snell's law, we get
$\mu  = {{\sin {i_1}} \over {\sin {r_1}}} = {{\sin A} \over {\sin (A/2)}}$
Using $\sin x = 2\sin {x \over 2}\cos {x \over 2}$, we get
$\mu  = {{2\sin A/2\cos A/2} \over {\sin A/2}} \Rightarrow \mu  = 2\cos {A \over 2}$ ....... (5)
Now, Eq. (4) becomes
$\sin {i_1} = \sin A\sqrt {{{\left( {2\cos {A \over 2}} \right)}^2} - 1 - \cos A} $
$ = \sin A\sqrt {4{{\cos }^2}{A \over 2} - 1}  - \cos A$
${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1}  - \cos A} \right]$
Hence, option (D) is correct.
$\bullet$ For option (B) : From Eq. (5), we have
$\mu  = 2\cos {A \over 2}$
$\cos {A \over 2} = {\mu  \over 2}$
$ \Rightarrow {A \over 2} = {\cos ^{ - 1}}\left( {{\mu  \over 2}} \right)$
$ \Rightarrow A = 2{\cos ^{ - 1}}\left( {{\mu  \over 2}} \right)$
Hence, option (B) is incorrect.
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Answer

For the angle of incidence ${i_1} = A,$ the ray inside the prism is parallel to the base of the prism

Answer

For this prism, the refractive index $\mu $ and the angle of prism $A$ are related as
$A = {1 \over 2}{\cos ^{ - 1}}\left( {{\mu  \over 2}} \right)$

Answer

At minimum deviation, the incident angle ${i_1}$ and the refracting angle ${r_1}$ at the first refracting surface are related by ${r_1} = \left( {{i_1}/2} \right)$

Answer

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is
${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1}  - \cos A} \right]$

Ray Optics

For an isosceles prism of angle A and refractive index \mu , it is found that the angle of minimum deviation {\delta _m} = A. Which of the following options is/are correct?

Choose the correct answer: