A lens of focal length 20 cm in air is made of glass with a refractive index of 1.6. What is its focal length when it is immersed in a liquid of refractive index 1.8?
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The focal length of a lens in a different medium can be calculated using the formula \( \frac{1}{f_{\text{medium}}} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} \right) \times \frac{1}{f_{\text{air}}} \).
The focal length of a lens in a medium is determined by the formula:
\[
\frac{1}{f_{\text{medium}}} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} \right) \times \frac{1}{f_{\text{air}}}
\]
where \( n_{\text{lens}} \) is the refractive index of the lens, \( n_{\text{medium}} \) is the refractive index of the medium, and \( f_{\text{air}} \) is the lens's focal length in air.
Given:
- \( f_{\text{air}} = 20 \, \text{cm} \)
- \( n_{\text{lens}} = 1.6 \)
- \( n_{\text{medium}} = 1.8 \)
Substituting these values into the formula yields:
\[
\frac{1}{f_{\text{medium}}} = \left( \frac{1.6}{1.8} \right) \times \frac{1}{20}
\]
Solving for \( f_{\text{medium}} \):
\[
f_{\text{medium}} = \frac{1}{\left( \frac{1.6}{1.8} \right) \times \frac{1}{20}} = -36 \, \text{cm}
\]
The focal length of the lens when submerged in the medium is \( -36 \, \text{cm} \).
The correct answer is (1) -36 cm.
