Question

A convex lens of refractive index \(1.5\) and focal length \(f=18\) cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is \( \alpha \times f \). Find the value of \( \alpha \). (Given: refractive index of water \(=\dfrac{4}{3}\))

Hint:

When a lens is immersed in a medium, always use the {relative refractive index} in the lens maker formula.

Answer 1

Correct Answer: 3

Concept: For a thin lens, the focal length depends on the refractive index of the lens material {relative to the surrounding medium}. The lens maker’s formula in a medium is: \[ \frac{1}{f_m}=\left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]
Step 1: Focal length of the lens in air In air, \( \mu_{\text{medium}}=1 \): \[ \frac{1}{f_{\text{air}}}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] Given: \[ f_{\text{air}}=18\text{ cm},\quad \mu=1.5 \] \[ \Rightarrow \left(\frac{1}{R_1}-\frac{1}{R_2}\right) =\frac{1}{18(1.5-1)}=\frac{1}{9} \]
Step 2: Focal length of the lens in water Relative refractive index: \[ \mu_{\text{rel}}=\frac{\mu_{\text{lens}}}{\mu_{\text{water}}} =\frac{1.5}{4/3}=\frac{9}{8} \] \[ \frac{1}{f_{\text{water}}} =\left(\frac{9}{8}-1\right)\frac{1}{9} =\frac{1}{8}\cdot\frac{1}{9} =\frac{1}{72} \] \[ \Rightarrow f_{\text{water}}=72\text{ cm} \]
Step 3: Difference in focal lengths \[ \Delta f=f_{\text{water}}-f_{\text{air}}=72-18=54\text{ cm} \]
Step 4: Express in the form \( \alpha f \) \[ \alpha=\frac{54}{18}=3 \] Final Answer: \[ \boxed{\alpha=3} \]