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 \(=1.5)\left(\sin \left(30^{\circ}\right)=0.5, \sin (48.6)=0.75\right)\)
- A \(17^{\circ}\)
- B \(12.6^{\circ}\)
- C \(16^{\circ}\)
- D \(18.6^{\circ}\)
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,
\(\delta=\left(i_1+i_2\right)-A=(0+48.6)-30=18.6^{\circ}\)
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,
\(\delta=\left(i_1+i_2\right)-A=(0+48.6)-30=18.6^{\circ}\)
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