For fixed values of radii of curvature of lens, power of the lens will be _________________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( P \propto (\mu - 1) \)
The power of a lens is \( P = 1/f \), and by the lens maker's formula
\( \displaystyle P = \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
For fixed \( R_1 \) and \( R_2 \) the bracket is a constant, so
\( \displaystyle P \propto (\mu - 1) \)
NCERT (2026-27):
Refraction by a Lens – “… So Eq. (9.20) can be written as
\( \displaystyle \frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \qquad (9.21) \)
Equation (9.21) is known as the lens maker’s formula.”
Power of a Lens – “Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it. … Thus,
\( \displaystyle P = \frac{1}{f} \qquad (9.25) \)”