Question

An optician prescribes a lens of power +2.5 D. The focal length of the lens in water is (Indices: Lens = 1.5, Water = 1.33):

• A. 40 cm
• B. 2660/17 cm
• C. 17/2660 cm
• D. 3000/17 cm
• E. 17/3000 cm

Hint:

Denser medium ⇒ weaker lens ⇒ longer focal length.

Answer 1

The Correct Option is B

Concept: Lens in a medium
When a lens is placed in a medium (like water), its power decreases because the refractive index difference between lens and medium reduces. Hence, focal length increases.

Step 1: Focal length in air
Given power: \[ P = 2.5 \text{ D} \] \[ f = \frac{1}{P} = \frac{1}{2.5} = 0.4 \text{ m} = 40 \text{ cm} \]

Step 2: Use lens-maker relation in medium
\[ f_m = f \times \frac{\mu_{\text{lens}} - 1}{\mu_{\text{lens}}/\mu_{\text{medium}} - 1} \]

Step 3: Substitute values
For glass: \[ \mu_{\text{lens}} = \frac{3}{2}, \quad \mu_{\text{water}} = \frac{4}{3} \] \[ f_m = 40 \times \frac{\frac{3}{2} - 1}{\frac{\frac{3}{2}}{\frac{4}{3}} - 1} \]

Step 4: Simplify step-by-step
\[ \frac{3}{2} - 1 = \frac{1}{2} \] \[ \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \] \[ \frac{9}{8} - 1 = \frac{1}{8} \] \[ f_m = 40 \times \frac{1/2}{1/8} = 40 \times 4 \] \[ f_m = 160 \text{ cm} \] Final Answer:
\[ \boxed{160 \text{ cm}} \] Conclusion:
Focal length increases in water, so the lens becomes weaker.