Question 4

The power $P$ of a lens of refractive index $n_g$ immersed in a medium of refractive index $n_L$ is given by the sum of the powers of its two surfaces:

$P = P_1 + P_2 = \frac{n_g - n_L}{R_1} + \frac{n_L - n_g}{R_2}$

$P = (n_g - n_L) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ Given:

Refractive index of glass, $n_g = 1.5$

Radii of curvature for a double convex lens, $R_1 = +20 \text{ cm} = +0.2 \text{ m}$ and $R_2 = -20 \text{ cm} = -0.2 \text{ m}$.

Substituting the values into the power formula:

$P = (1.5 - n_L) \left( \frac{1}{0.2} - \frac{1}{-0.2} \right)$

$P = (1.5 - n_L) \left( 5 + 5 \right)$

$P = 10(1.5 - n_L)$

$P = 15 - 10n_L$

This is a linear equation of the form $y = mx + c$, where the slope $m = -10$ and the intercept $c = 15$.

Let's check specific points:

• For $n_L = 1.0$: $P = 15 - 10(1.0) = 5 \text{ D}$

• For $n_L = 1.5$: $P = 15 - 10(1.5) = 0 \text{ D}$

• For $n_L = 2.0$: $P = 15 - 10(2.0) = -5 \text{ D}$

The graph that represents this linear variation is shown in option (B).