Solveeit Logo

Question

Question: The angle of deviation when light is incident at an angle of \(45{}^\circ \) on one of the refractin...

The angle of deviation when light is incident at an angle of 4545{}^\circ on one of the refracting faces of an equilateral prism of refractive index 1.414 is
A. 40 B. 30 C. 45 D. 50 \begin{aligned} & \text{A}\text{. 40}{}^\circ \\\ & \text{B}\text{. 30}{}^\circ \\\ & \text{C}\text{. 45}{}^\circ \\\ & \text{D}\text{. 50}{}^\circ \\\ \end{aligned}

Explanation

Solution

Hint: By using the formula of refractive index for prism, we get the correct answer. The angle of deviation of a ray of light passing through a prism depends upon its material as well as on the angle of incidence. Here it means it depends upon the refractive index of a material of prism.

Complete answer step by step:
The given prism is equilateral so the three angles are of 6060{}^\circ . So, A=60\angle A=60{}^\circ .

The angle of incident is given,

i=45\angle i=45{}^\circ

We know that the equation for refractive index of prism,

μ=sin[A+δm2]sin(A2)\mu =\dfrac{\sin \left[ \dfrac{A+{{\delta }_{m}}}{2} \right]}{\sin \left( \dfrac{A}{2} \right)}
where µ is the refractive index of prism and δm{{\delta }_{m}} is angle of deviation.
Substituting given values in above equation we get,

1.414=sin[60+δm2]sin(602)=sin[60+δm2]sin(30)1.414=\dfrac{\sin \left[ \dfrac{60{}^\circ +{{\delta }_{m}}}{2} \right]}{\sin \left( \dfrac{60{}^\circ }{2} \right)}=\dfrac{\sin \left[ \dfrac{60{}^\circ +{{\delta }_{m}}}{2} \right]}{\sin \left( 30{}^\circ \right)}

We know sin(30)=12=0.5\sin \left( 30{}^\circ \right)=\dfrac{1}{2}=0.5, so above equation can be written as,

1.414×0.5=sin[60+δm2]1.414\times 0.5=\sin \left[ \dfrac{60{}^\circ +{{\delta }_{m}}}{2} \right]
0.707=sin[60+δm2]0.707=\sin \left[ \dfrac{60{}^\circ +{{\delta }_{m}}}{2} \right]

Multiplying both sides by sin1{{\sin }^{-1}}, we get

sin1(0.707)=[60+δm2]{{\sin }^{-1}}(0.707)=\left[ \dfrac{60+{{\delta }_{m}}}{2} \right]
We know, sin(45)=12=0.707\sin \left( 45{}^\circ \right)=\dfrac{1}{\sqrt{2}}=0.707, so above equation can be written as,

& 45{}^\circ =\dfrac{60{}^\circ +{{\delta }_{m}}}{2} \\\ & \Rightarrow {{\delta }_{m}}=45{}^\circ \times 2-60{}^\circ \\\ \end{aligned}$$ $${{\delta }_{m}}={{30}^{0}}$$ Therefore, the angle of deviation when light is incident at an angle of $45{}^\circ $ on one of the refracting faces of an equilateral prism of refractive index 1.414 is $${{\delta }_{m}}={{30}^{0}}$$ Hence the correct option is B. Note: The given prism is equilateral, so the angle of the prism is known and also the angle of incidence is given with the refractive index of the prism. Here, Snell’s equation is used and substituted values of ‘i' and ‘r’. And they are $$i=\dfrac{A+{{\delta }_{m}}}{2}$$ and $$r=\dfrac{A}{2}$$. So, if we know which material is used and what is the angle of incident, we can solve any problem of prism.