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Question: For the angle of minimum deviation of a prism is equal to its refracting angle, the prism must be ma...

For the angle of minimum deviation of a prism is equal to its refracting angle, the prism must be made of a material whose refractive index:
(A) Between 2 and 2\sqrt{2}
(B) Is less than 1
(C) Is greater than 2
(D) Lies between 2\sqrt{2} and 1

Explanation

Solution

Hint : We can find the refractive index corresponding to minimum deviation by using a prism formula.
n=sin(A+Sm)2sinA2n=\dfrac{\sin \dfrac{(A+{{S}_{m}})}{2}}{\sin \dfrac{A}{2}}
Here, n is refractive index of material,
A is the angle of the prism.
Sm{{S}_{m}} is the angle of minimum deviation
Use the condition, refractive index is minimum when angle of prism is 90\text{9}0{}^\circ
Refractive index is maximum when the angle of the prism is 00{}^\circ .

Complete step by step answer
The prism formula is given by
n=sin(A+Sm)2sinA2n=\dfrac{\sin \dfrac{(A+{{S}_{m}})}{2}}{\sin \dfrac{A}{2}}
Here n is a refractive index.
We know deviation will be minimum, angle of incidence is equal to angle of prism.
Sm=2iA{{S}_{m}}=2i-A
Sm=2AA=A{{S}_{m}}=\text{2}AA=A
From prism formula, put the value of minimum deviation,
n = n=sin(A+A)2sinA2=sin AsinA2n=\dfrac{\sin \dfrac{(A+A)}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin \text{ }A}{\sin \dfrac{A}{2}}
Use[sin A=2 sinA2cos A2]\left[ \text{sin A=2 sin}\dfrac{A}{2}\text{cos }\dfrac{A}{2} \right]
The refractive index is given by,
n=2SinA2CosA2SinA2n=\dfrac{2\operatorname{Sin}\dfrac{A}{2}\operatorname{Cos}\dfrac{A}{2}}{\operatorname{Sin}\dfrac{A}{2}}
n=2CosA2n=2\operatorname{Cos}\dfrac{A}{2}
This is the formula for prism,
Now, we have to find a minimum refractive index.
Angle of prism A varies from 00{}^\circ to 90\text{9}0{}^\circ
For nmin{{n}_{\min }}, minimum value of refractive index is given by
Put A=90\text{A}=\text{9}0{}^\circ
n=2Cos902n=2\operatorname{Cos}\dfrac{90{}^\circ }{2} Use[Cos 45=12]\left[ \text{Cos 45}{}^\circ =\dfrac{1}{\sqrt{2}} \right]
= 2Cos452\operatorname{Cos}45{}^\circ
nmin=2×12 = 2{{n}_{\min }}=2\times \dfrac{1}{\sqrt{2}}\text{ = }\sqrt{2}
Fornmax{{n}_{\max }}, maximum value of refractive index
A= 0°
n=2 Cos 0=2\text{n}=\text{2 Cos }0{}^\circ =\text{2} (1)  [Cos 0= 1]~[\text{Cos }0{}^\circ =\text{ 1}]
nmax=2{{n}_{\max }}=2
Hence, the value of refractive index varies between 2 and 2\sqrt{2}

Note
When refracting angle is small, the deviation is calculated from,
S =(n1)\left( n-1 \right)A
When refraction angle is bigger for the prism the deviation is calculated from, S = (i1+i2)A({{i}_{1}}+{{i}_{2}})-Awhere i1{{i}_{1}}and i2{{i}_{2}}are angle of incidence on different faces of prism.