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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2022
  • Hard
  • Numerical

The \(X-Y\) plane be taken as the boundary between two transparent media \(M _{1}\) and \(M _{2} . M _{1}\) in \(Z \geq 0\) has a refractive index of \(\sqrt{2}\) and \(M _{2}\) with \(Z <0\) has a refractive index of \(\sqrt{3}\). A ray of light travelling in \(M_{1}\) along the direction given by the vector \(\overrightarrow{ A }=4 \sqrt{3 i }-3 \sqrt{3 \hat{j}}-5 \hat{ k }\), is incident on the plane of separation. The value of difference between the angle of incident in \(M _{1}\) and the angle of refraction in \(M _{2}\) will be \(....\) degree.

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Answer & Solution

Correct Answer

(B) \(15\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ A }=4 \sqrt{3 \hat{i}}-3 \sqrt{3 \hat{j}}-5 \hat{ k }\)

As incident vector \(A\) makes \(i\) angle with normal \(z-\)axis refracted vector \(R\) makes \(r\) angle with normal \(z\) - axis with help of direction cosine

\(i=\cos ^{-1}\left(\frac{A_{z}}{A}\right)=\cos ^{-1}\left(\frac{5}{\sqrt{(4 \sqrt{3})^{2}+(3 \sqrt{3})^{2}+5^{2}}}\right)\)

\(=\cos ^{-1}\left(\frac{5}{10}\right) \Rightarrow i=60^{\circ}\)

\(\sqrt{2} \sin 60=\sqrt{3} \times \sin r\)

\(r=45^{\circ}\)

Difference between \(i \) & \(r =60-45=15\)

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