Question

When a convex lens is dipped in water, its focal length and nature become:

• A. increased, convex lens
• B. decreased, convex lens
• C. increased, concave lens
• D. decreased, concave lens

Hint:

Use the lens maker's formula: in water the factor \( (\mu_g/\mu_w - 1) \) is smaller but still positive, so \( f \) grows while the lens stays converging.

Answer 1

The Correct Option is A

Step 1: Write the lens maker's formula.
\[ \frac{1}{f} = \left(\,{}^{a}\mu_g - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right), \] where \( {}^{a}\mu_g \) is the refractive index of the lens material relative to the surrounding medium.

Step 2: Compare air and water as the medium.
In air, the relevant factor is \( ({}^{a}\mu_g - 1) \) with \( {}^{a}\mu_g \approx 1.5 \), giving \( 1.5 - 1 = 0.5 \).
In water, the factor becomes \( ({}^{w}\mu_g - 1) \) where \[ {}^{w}\mu_g = \frac{{}^{a}\mu_g}{{}^{a}\mu_w} = \frac{1.5}{1.33} \approx 1.13, \] giving \( 1.13 - 1 = 0.13 \).

Step 3: See the effect on \( f \).
The bracket \( \left(\frac{1}{R_1}-\frac{1}{R_2}\right) \) is unchanged. Since the leading factor drops from 0.5 to about 0.13, \( 1/f \) becomes much smaller, so \( f \) increases (roughly \( 0.5/0.13 \approx 4 \) times).

Step 4: Check the nature.
The factor \( ({}^{w}\mu_g - 1) \) is still positive because glass is denser than water. A positive factor keeps \( f \) positive, so the lens stays converging (convex).

Step 5: Conclude.
The focal length increases and the lens remains convex, so option (i) is correct. (It would flip to diverging only if the medium were optically denser than glass.)
\[\boxed{f \uparrow,\ \text{still convex}}\]