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.

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: