TS EAMCET · Physics · Ray Optics
A prism of refractive index \(\mu\) and angle \(A\) is placed in the minimum deviation position. If the angle of minimum deviation is \(A\), then the value of \(A\) in terms of \(\mu\) is :
- A \(\sin ^{-1}\left(\frac{\mu}{2}\right)\)
- B \(\sin ^{-1} \sqrt{\frac{\mu-1}{2}}\)
- C \(2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
- D \(\cos ^{-1}\left(\frac{\mu}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
Step-by-step Solution
Detailed explanation
Refractive index of material of prism \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\) but \(\quad \delta_m=A\) \(\mu=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}\) \(\mu=\frac{\sin A}{\sin A / 2}\) \(\mu=\frac{2 \sin A / 2 \cos A / 2}{\sin A / 2}\)…
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