Optics — Medium JEE physics MCQ
A double convex lens made of glass (refractive index $1.5$) has both radii of curvature equal to $20$ cm. The power of the lens is:
- A. $2.5$ D
- B. $5$ D
- C. $10$ D
- D. $1.25$ D
Solution
Using the lens maker's equation:
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
For a double convex lens: $R_1 = +20$ cm, $R_2 = -20$ cm
$$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right)$$
$$\frac{1}{f} = 0.5 \times \frac{2}{20} = 0.5 \times \frac{1}{10} = \frac{1}{20} \text{ cm}^{-1}$$
So $f = 20$ cm $= 0.2$ m
Power: $P = \frac{1}{f} = \frac{1}{0.2} = 5$ D