Question

Figure I given below shows a glass vessel, partially filled with water. A narrow beam of light is incident vertically down into the water and passes straight through. Figure II shows the vessel glass tilted until the angle $\theta$, such that the light is refracted along the lower surface of the glass. If refractive indices of air, water and glass are $1,\frac{4}{3}$ and 1.5 respectively, then ,

• A. $sin\theta =\frac{3}{4}$
• B. $cos\theta =\frac{3}{4}$
• C. $sin\theta =\frac{8}{9}$
• D. $cos\theta =\frac{8}{9}$

Answer 1

The correct option is A $sin\theta =\frac{3}{4}$

We have

$\mu =\frac{1}{sinC}$

For the refraction at glass-air interface, the ray passing through the point $Q$ is given by

$\frac{sinr}{sin{90}^o}=\frac{1}{1.5}$

$⟹sinr=\frac{1}{1.5}$

Now for the refraction at water-glass interface, the ray passing through point P is

$\frac{sin\theta }{sinr}=\frac{1.5}{\frac{4}{3}}$

$⟹sin\theta =\frac{3}{4}$