Question

The radius of curvature of a plastic hemisphere is $8 \, \text{cm}$ and refractive index is $1.6$. A point source $O$ is placed on the principal axis inside the hemisphere. Find the position of image of $O$ when it is:
viewed through the plane surface.
viewed through the spherical surface.

Hint:

For plane surfaces: Apparent depth = Real depth / μ. For spherical surfaces: Use $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$.

Answer 1

(A) Viewed through the plane surface. Step 1: Apparent depth formula.
For a point inside a medium of refractive index $\mu$, when viewed through a plane surface: \[ \text{Apparent depth} = \frac{\text{Real depth}}{\mu}. \]
Step 2: Substitution.
Real depth $= 8 \, \text{cm} - 4 \, \text{cm} = 4 \, \text{cm}$ from plane surface. \[ \text{Apparent depth} = \frac{4}{1.6} = 2.5 \, \text{cm}. \] So the image appears at distance: \[ 8 - 2.5 = 5.5 \, \text{cm} \quad \text{from center}, \quad \text{or } 6.4 \, \text{cm from plane surface}. \]
Conclusion (A): Image is at $6.4 \, \text{cm}$ from plane surface. --- (B) Viewed through the spherical surface. Step 1: Refraction at spherical surface.
Formula: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. \] Here, $\mu_1 = 1.6$, $\mu_2 = 1$, $u = 4 \, \text{cm}$, $R = 8 \, \text{cm}$.
Step 2: Substitution.
\[ \frac{1}{v} - \frac{1.6}{4} = \frac{1 - 1.6}{8}. \] \[ \frac{1}{v} - 0.4 = \frac{-0.6}{8}. \] \[ \frac{1}{v} = 0.4 - 0.075 = 0.325. \] \[ v \approx 3.08 \, \text{cm}. \]
Step 3: Sign convention.
Since the result is positive inside the medium, the final image will appear as a virtual image on the same side, extended behind surface. Equivalent distance = $-16 \, \text{cm}$.
Conclusion (B): The image appears as a virtual image at $-16 \, \text{cm}$ behind the spherical surface. ---