Question

An equiconvex lens of refractive index $\frac{3}{2}$ is combined with another equi concave lens of refractive index $\frac{4}{3}$ as shown. If radius of curvature of each surface is 20cm, the equivalent focal length of the system is

Answer 1

60 cm

Answer 2

To find the equivalent focal length of the system, we need to calculate the focal length of each lens individually using the lens maker's formula and then combine them using the formula for lenses in contact.

The lens maker's formula is: $\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Given radius of curvature for each surface is $R = 20$ cm. We follow the sign convention where light travels from left to right.

1. Equiconvex Lens (Lens 1):

• Refractive index $\mu_1 = \frac{3}{2}$
• For an equiconvex lens, the first surface (left) is convex, so its center of curvature is on the right: $R_1 = +20$ cm.
• The second surface (right) is convex, so its center of curvature is on the left: $R_2 = -20$ cm.

Applying the lens maker's formula for Lens 1: $\frac{1}{f_1} = \left(\frac{3}{2} - 1\right) \left(\frac{1}{+20} - \frac{1}{-20}\right)$ $\frac{1}{f_1} = \left(\frac{1}{2}\right) \left(\frac{1}{20} + \frac{1}{20}\right)$ $\frac{1}{f_1} = \left(\frac{1}{2}\right) \left(\frac{2}{20}\right)$ $\frac{1}{f_1} = \frac{1}{20}$ $f_1 = +20$ cm

2. Equiconcave Lens (Lens 2):

• Refractive index $\mu_2 = \frac{4}{3}$
• For an equiconcave lens, the first surface (left) is concave, so its center of curvature is on the left: $R_1 = -20$ cm.
• The second surface (right) is concave, so its center of curvature is on the right: $R_2 = +20$ cm.

Applying the lens maker's formula for Lens 2: $\frac{1}{f_2} = \left(\frac{4}{3} - 1\right) \left(\frac{1}{-20} - \frac{1}{+20}\right)$ $\frac{1}{f_2} = \left(\frac{1}{3}\right) \left(-\frac{1}{20} - \frac{1}{20}\right)$ $\frac{1}{f_2} = \left(\frac{1}{3}\right) \left(-\frac{2}{20}\right)$ $\frac{1}{f_2} = \left(\frac{1}{3}\right) \left(-\frac{1}{10}\right)$ $\frac{1}{f_2} = -\frac{1}{30}$ $f_2 = -30$ cm

3. Equivalent Focal Length of the System: Since the two lenses are combined in contact, the equivalent focal length ($F_{eq}$) is given by: $\frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$ $\frac{1}{F_{eq}} = \frac{1}{20} + \frac{1}{-30}$ $\frac{1}{F_{eq}} = \frac{1}{20} - \frac{1}{30}$

To combine these fractions, find a common denominator, which is 60: $\frac{1}{F_{eq}} = \frac{3}{60} - \frac{2}{60}$ $\frac{1}{F_{eq}} = \frac{1}{60}$ $F_{eq} = +60$ cm

The equivalent focal length of the system is 60 cm.