Question

Calculate the focal length of a biconvex lens if the radii of its surfaces are 50 cm and 20 cm, and index of refraction of the lens glass = 1.2.

(a) 0.014 cm

(b) 0.715 cm

(c) 0.14 cm

(d) 71.5 cm

I got this question in examination.

Origin of the question is Ray Optics topic in section Ray Optics and Optical Instruments of Physics – Class 12

Answer 1

Right answer is (d) 71.5 cm

The best I can explain: Given: R1 = 50 cm; R2 = 20 cm; μ2 = 1.2

Required equation ➔ \(\frac {1}{f}\)=(μ-1)\( ( \frac {1}{R_1} – \frac {1}{R_2} ) \)

 \(\frac {1}{f}=( \frac {\mu 2}{\mu 1} -1 ) ( \frac {1}{R_1} – \frac {1}{R_2} ) \)

 \(\frac {1}{f}=( \frac {1.2}{1} -1 ) ( \frac {1}{50} – \frac {1}{-20} ) \)

 \(\frac {1}{f}\)=(0.2)\( ( \frac {1}{50} + \frac {1}{20} ) \)

 \(\frac {1}{f}\)=(0.2)\( ( \frac {20}{1000} + \frac {50}{1000} ) \)

 \(\frac {1}{f}\)=(0.2)\( ( \frac {7}{100} ) \)

\(\frac {1}{f}\)=0.014

  f=+71.5 cm