Question

If the radius of curvature of each face of the biconcave lens, made of glass of refractive index 1.25 is 50 cm, then what is the focal length of the lens in air?

(a) 30 cm

(b) -30 cm

(c) 5 cm

(d) -5 cm

I have been asked this question in an interview.

I would like to ask this question from Reflection of Light by Spherical Mirrors topic in portion Ray Optics and Optical Instruments of Physics – Class 12

Answer 1

Correct choice is (b) -30 cm

The best explanation: \(\frac {1}{f}\) = (μ-1)\( [ \frac {1}{R_1} – \frac {1}{R_2} ] \)

\(\frac {1}{f}\) = (1.25 – 1)\( [ \frac {1}{-50} – \frac {1}{50} ] \)

\(\frac {1}{f} = \frac {-1}{100}\)

f = -100 cm

Therefore, the focal length of the lens in air is -100 cm.