माना कि एक सघन माध्यम का एक विरल माध्यम के सापेक्ष अपवर्तनांक \(n _{12}\) है तथा उसका क्रान्तिक कोण \(\theta_{C}\) है। जब प्रकाश एक आपतन कोण \(A\) से सघन से विरल माध्यम में जाता है तो उसका एक भाग परावर्तित होता है और बचा हुआ भाग अपवर्तित होता है। परावर्तित और अपवर्तित किरणों के बीच कोण \(90^{\circ}\) है। कोण \(A\) का मान होगा

A \(\frac{1}{{{{\cos }^{ - 1}}\,\left( {\sin {\mkern 1mu} {\theta _C}} \right)}}\)
B \(\frac{1}{{{{\tan }^{ - 1}}\,\left( {\sin {\mkern 1mu} {\theta _C}} \right)}}\)
C \({{{\cos }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)
D \({{{\tan }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \({{{\tan }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)

Step-by-step Solution

Detailed explanation

From Snell's law, \(\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{D}}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) ..... \((i)\) \(\because \,\angle {\text{i}} = {\text{A}}\) and \(\angle {\text{r}} = \left( {{{90}^o} - {\text{A}}} \right)\) We also know that,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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