Solution to Question 13
1. Optical Path Analysis
The lens is plano-convex with the curved side up and partially immersed in water. Since there is no reflection, we only consider refraction. The problem states that two images are formed at different depths $d$ and $D$. This implies two distinct optical regions created by the water level:
- Region 1 (Central Circle): Light passes through Air $\to$ Glass (Curved Surface) $\to$ Water (Flat Surface).
- Region 2 (Outer Annulus): Light passes through Water $\to$ Glass (Curved Surface) $\to$ Water (Flat Surface).
2. Calculating Focal Depths
Let $R$ be the radius of curvature of the curved surface, $\mu$ be the refractive index of the lens, and $\mu_0$ be the refractive index of water.
Path 1 (Central part): Air $\to$ Glass $\to$ Water
Refraction at curved surface (Air to Glass): $v’ = \frac{\mu R}{\mu – 1}$.
Refraction at flat surface (Glass to Water): $\frac{\mu_0}{f_1} = \frac{\mu}{v’} \implies f_1 = v’ \frac{\mu_0}{\mu}$.
$$ f_1 = \frac{\mu R}{\mu – 1} \cdot \frac{\mu_0}{\mu} = \frac{\mu_0 R}{\mu – 1} $$
This corresponds to the smaller depth $d$.
$$ d = \frac{\mu_0 R}{\mu – 1} \quad \text{…(i)} $$
Path 2 (Outer part): Water $\to$ Glass $\to$ Water
Refraction at curved surface (Water to Glass): $v” = \frac{\mu R}{\mu – \mu_0}$.
Refraction at flat surface (Glass to Water): $f_2 = v” \frac{\mu_0}{\mu}$.
$$ f_2 = \frac{\mu R}{\mu – \mu_0} \cdot \frac{\mu_0}{\mu} = \frac{\mu_0 R}{\mu – \mu_0} $$
This corresponds to the larger depth $D$.
$$ D = \frac{\mu_0 R}{\mu – \mu_0} \quad \text{…(ii)} $$
3. Solving for $\mu$ and $R$
From (i) and (ii), we can isolate $\mu_0 R$:
$$ d(\mu – 1) = \mu_0 R \quad \text{and} \quad D(\mu – \mu_0) = \mu_0 R $$
Equating the two expressions:
$$ d\mu – d = D\mu – D\mu_0 $$
$$ D\mu_0 – d = \mu(D – d) $$
$$ \mu = \frac{D\mu_0 – d}{D – d} $$
Substituting $\mu$ back into (i) to find $R$:
$$ R = \frac{d(\mu – 1)}{\mu_0} = \frac{d}{\mu_0} \left( \frac{D\mu_0 – d}{D – d} – 1 \right) $$
$$ R = \frac{d}{\mu_0} \left( \frac{D\mu_0 – d – D + d}{D – d} \right) = \frac{d}{\mu_0} \frac{D(\mu_0 – 1)}{D – d} $$
$$ R = \frac{dD(\mu_0 – 1)}{\mu_0(D – d)} $$
4. Finding Depth $h$
The problem states the images are “equally bright.” Brightness is proportional to the light-gathering area (aperture area) of the respective lens regions.
Let $A$ be the total aperture radius of the lens, and $r$ be the radius of the circular boundary where the water level cuts the curved surface.
Area of Central Region = $\pi r^2$
Area of Outer Region = $\pi A^2 – \pi r^2$
For equal brightness:
$$ \pi r^2 = \pi A^2 – \pi r^2 \implies 2\pi r^2 = \pi A^2 \implies r^2 = \frac{A^2}{2} $$
For a spherical cap of thickness (sag) $y$, the radius $r$ is related by $r^2 \approx 2Ry$ (for thin lenses).
Total thickness is $H$, so $A^2 \approx 2RH$.
The top part (in air) has thickness $H – h$. So $r^2 \approx 2R(H – h)$.
Substituting these into the area relation:
$$ 2R(H – h) = \frac{1}{2} (2RH) $$
$$ 2(H – h) = H \implies 2H – 2h = H \implies h = \frac{H}{2} $$
$$ R = \frac{dD(\mu_0 – 1)}{\mu_0(D – d)}, \quad \mu = \frac{D\mu_0 – d}{D – d}, \quad h = \frac{H}{2} $$
1. Optical Path Analysis
The lens is plano-convex with the curved side up and partially immersed in water. Since there is no reflection, we only consider refraction. The problem states that two images are formed at different depths $d$ and $D$. This implies two distinct optical regions created by the water level:
- Region 1 (Central Circle): Light passes through Air $\to$ Glass (Curved Surface) $\to$ Water (Flat Surface).
- Region 2 (Outer Annulus): Light passes through Water $\to$ Glass (Curved Surface) $\to$ Water (Flat Surface).
2. Calculating Focal Depths
Let $R$ be the radius of curvature of the curved surface, $\mu$ be the refractive index of the lens, and $\mu_0$ be the refractive index of water.
Path 1 (Central part): Air $\to$ Glass $\to$ Water
Refraction at curved surface (Air to Glass): $v’ = \frac{\mu R}{\mu – 1}$.
Refraction at flat surface (Glass to Water): $\frac{\mu_0}{f_1} = \frac{\mu}{v’} \implies f_1 = v’ \frac{\mu_0}{\mu}$.
$$ f_1 = \frac{\mu R}{\mu – 1} \cdot \frac{\mu_0}{\mu} = \frac{\mu_0 R}{\mu – 1} $$
This corresponds to the smaller depth $d$.
$$ d = \frac{\mu_0 R}{\mu – 1} \quad \text{…(i)} $$
Path 2 (Outer part): Water $\to$ Glass $\to$ Water
Refraction at curved surface (Water to Glass): $v” = \frac{\mu R}{\mu – \mu_0}$.
Refraction at flat surface (Glass to Water): $f_2 = v” \frac{\mu_0}{\mu}$.
$$ f_2 = \frac{\mu R}{\mu – \mu_0} \cdot \frac{\mu_0}{\mu} = \frac{\mu_0 R}{\mu – \mu_0} $$
This corresponds to the larger depth $D$.
$$ D = \frac{\mu_0 R}{\mu – \mu_0} \quad \text{…(ii)} $$
3. Solving for $\mu$ and $R$
From (i) and (ii), we can isolate $\mu_0 R$: $$ d(\mu – 1) = \mu_0 R \quad \text{and} \quad D(\mu – \mu_0) = \mu_0 R $$ Equating the two expressions: $$ d\mu – d = D\mu – D\mu_0 $$ $$ D\mu_0 – d = \mu(D – d) $$ $$ \mu = \frac{D\mu_0 – d}{D – d} $$
Substituting $\mu$ back into (i) to find $R$: $$ R = \frac{d(\mu – 1)}{\mu_0} = \frac{d}{\mu_0} \left( \frac{D\mu_0 – d}{D – d} – 1 \right) $$ $$ R = \frac{d}{\mu_0} \left( \frac{D\mu_0 – d – D + d}{D – d} \right) = \frac{d}{\mu_0} \frac{D(\mu_0 – 1)}{D – d} $$ $$ R = \frac{dD(\mu_0 – 1)}{\mu_0(D – d)} $$
4. Finding Depth $h$
The problem states the images are “equally bright.” Brightness is proportional to the light-gathering area (aperture area) of the respective lens regions.
Let $A$ be the total aperture radius of the lens, and $r$ be the radius of the circular boundary where the water level cuts the curved surface.
Area of Central Region = $\pi r^2$
Area of Outer Region = $\pi A^2 – \pi r^2$
For equal brightness:
$$ \pi r^2 = \pi A^2 – \pi r^2 \implies 2\pi r^2 = \pi A^2 \implies r^2 = \frac{A^2}{2} $$
For a spherical cap of thickness (sag) $y$, the radius $r$ is related by $r^2 \approx 2Ry$ (for thin lenses).
Total thickness is $H$, so $A^2 \approx 2RH$.
The top part (in air) has thickness $H – h$. So $r^2 \approx 2R(H – h)$.
Substituting these into the area relation:
$$ 2R(H – h) = \frac{1}{2} (2RH) $$
$$ 2(H – h) = H \implies 2H – 2h = H \implies h = \frac{H}{2} $$