Question

A circular beam of light having diameter 'd' falls on a plane liquid surface from air. The angle of incidence is \(45^{\circ }\) and refractive index of liquid is 'n'. The diameter of the refracted beam is
\(\left(\cos 45^{0}=1/\sqrt{2},\ \sin 45^{0}=1/\sqrt{2}\right)\)

• A. \((2n^{2}-1)d\)
• B. \(\frac{\sqrt{2n^{2}-1}}{n}\ d\)
• C. \(d\)
• D. \(\frac{\sqrt{n^{2}-1}}{n}\ d\)

Hint:

The beam widths are projections of the lit strip on the surface, and Snell's law gives the angle of refraction.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The beam widths are projections of the strip the beam lights on the surface. Snell's law links the angles of incidence and refraction.

Step 2: Beam Widths
Let the lit strip have length \(L\). The incident diameter is \(d=L\cos 45^{\circ}\) and the refracted diameter is \(d'=L\cos r\). So \(d'=d\,\dfrac{\cos r}{\cos 45^{\circ}}\).

Step 3: Angle of Refraction
Snell's law gives \(\sin r=\dfrac{\sin 45^{\circ}}{n}=\dfrac{1}{\sqrt{2}\,n}\). Hence \(\cos r=\sqrt{1-\dfrac{1}{2n^{2}}}=\dfrac{\sqrt{2n^{2}-1}}{\sqrt{2}\,n}\).

Step 4: Calculation
\(d'=\sqrt{2}\,d\cdot\dfrac{\sqrt{2n^{2}-1}}{\sqrt{2}\,n}=\dfrac{\sqrt{2n^{2}-1}}{n}\,d\). For \(n=1\) this gives \(d'=d\), as it should.

Final Answer:
The refracted beam has diameter \(\dfrac{\sqrt{2n^{2}-1}}{n}\,d\). \[ \boxed{\text{(B) }\dfrac{\sqrt{2n^{2}-1}}{n}\,d} \]