Question:
A spherical glass bottle having negligible wall thickness is placed in air. When the bottle is completely filled with water, its focal length is $f$. If the water is replaced by another transparent liquid of higher refractive index, then the focal length changes to $f'$. Then
A spherical glass bottle having negligible wall thickness is placed in air. When the bottle is completely filled with water, its focal length is $f$. If the water is replaced by another transparent liquid of higher refractive index, then the focal length changes to $f'$. Then
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A simple mnemonic: Higher refractive index $\implies$ stronger bending $\implies$ shorter focal length.
This is always true for any positive (converging) lens.
This is always true for any positive (converging) lens.
Updated On: Jun 16, 2026
- $f' \lt f$
- $f' \gt f$
- $f' = f = \infty$
- $f' = f$, but finite
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem presents a spherical bottle filled with liquid acting as a thick spherical lens in air.
We need to determine how the focal length changes when the refractive index of the liquid inside is increased.
Step 2: Key Formula or Approach:
For a sphere of radius $R$ and refractive index $n$ placed in a medium of refractive index $n_0 = 1$ (air), the focal length measured from the center is given by the thick lens formula or refraction at spherical surfaces:
Alternatively, we can analyze the bending power of the sphere:
- A medium with a higher refractive index causes greater bending of light rays at the interfaces.
- More bending means the rays converge closer to the sphere, resulting in a smaller focal length.
Step 3: Detailed Explanation:
• When light enters the spherical bottle filled with a liquid of refractive index $n_1$ from air ($n_0 = 1$), it undergoes refraction at the front and back spherical surfaces.
• The optical power $P$ of a lens represents its ability to converge or diverge light rays, which is inversely proportional to its focal length:
\[ P = \frac{1}{f} \]
• The power of a refracting surface is directly proportional to the difference in refractive index across the interface ($\Delta n = n_{\text{liquid}} - n_{\text{air}}$).
• Since the second liquid has a higher refractive index:
\[ n_{\text{liquid}}' \gt n_{\text{liquid}} \]
- The difference in refractive index increases:
\[ (n_{\text{liquid}}' - 1) \gt (n_{\text{liquid}} - 1) \]
- This increases the converging power of both the front and back surfaces of the spherical lens:
\[ P' \gt P \]
• Since $P = \frac{1}{f}$, an increase in power results in a decrease in focal length:
\[ f' \lt f \]
Step 4: Final Answer:
The correct option is $f' \lt f$.
The problem presents a spherical bottle filled with liquid acting as a thick spherical lens in air.
We need to determine how the focal length changes when the refractive index of the liquid inside is increased.
Step 2: Key Formula or Approach:
For a sphere of radius $R$ and refractive index $n$ placed in a medium of refractive index $n_0 = 1$ (air), the focal length measured from the center is given by the thick lens formula or refraction at spherical surfaces:
Alternatively, we can analyze the bending power of the sphere:
- A medium with a higher refractive index causes greater bending of light rays at the interfaces.
- More bending means the rays converge closer to the sphere, resulting in a smaller focal length.
Step 3: Detailed Explanation:
• When light enters the spherical bottle filled with a liquid of refractive index $n_1$ from air ($n_0 = 1$), it undergoes refraction at the front and back spherical surfaces.
• The optical power $P$ of a lens represents its ability to converge or diverge light rays, which is inversely proportional to its focal length:
\[ P = \frac{1}{f} \]
• The power of a refracting surface is directly proportional to the difference in refractive index across the interface ($\Delta n = n_{\text{liquid}} - n_{\text{air}}$).
• Since the second liquid has a higher refractive index:
\[ n_{\text{liquid}}' \gt n_{\text{liquid}} \]
- The difference in refractive index increases:
\[ (n_{\text{liquid}}' - 1) \gt (n_{\text{liquid}} - 1) \]
- This increases the converging power of both the front and back surfaces of the spherical lens:
\[ P' \gt P \]
• Since $P = \frac{1}{f}$, an increase in power results in a decrease in focal length:
\[ f' \lt f \]
Step 4: Final Answer:
The correct option is $f' \lt f$.
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