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MHT CET · Physics · Ray Optics

The refracting angle of a glass prism is \(30^{\circ}\). A ray is incident on one of the faces and is perpendicular to it. The angle of deviation \(\delta\) between the incident ray and that leaves the prism is
(Refractive index of glass \(=1.5)\left(\sin \left(30^{\circ}\right)=0.5, \sin (48.6)=0.75\right)\)

  1. A \(17^{\circ}\)
  2. B \(12.6^{\circ}\)
  3. C \(16^{\circ}\)
  4. D \(18.6^{\circ}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(18.6^{\circ}\)

Step-by-step Solution

Detailed explanation

Given, \(A=30^{\circ}, \mu=1.5\) and \(i_1=0^{\circ}\)
Since, \(i_0\), therefore, \(r_1\) is also equal to \(0^{\circ}\)
Further, since, \(r_1+r_2=A\)
\(\therefore r_2=A=30^{\circ}\)
Using Snell's Law, \(\mu=\frac{\sin i_2}{\sin r_2}\)
\(\Rightarrow 1.5=\frac{\sin i_2}{\sin 30^{\circ}}\)
or \(i_2=1.5 \sin 30^{\circ}=1.5 \times \frac{1}{2}=0.75\)
or \(i_2=1.5 \sin 30^{\circ}=1.5 \times \frac{1}{2}=0.75\)
Now, the deviation,
\(\delta=\left(i_1+i_2\right)-A=(0+48.6)-30=18.6^{\circ}\)