The radius of curvature of each surface of a biconvex lens is 30 cm and the refractive index of the material of the lens is 1.5. An object is placed at 75 cm from the lens, the image is formed at distance x from the lens. In a situation when the refractive index of the material of lens is 1.6, the image is formed for the same object at distance y from the lens. The difference between x and y is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12.5 cm
The focal length follows from the lens maker’s formula and the image distance from the lens formula, with the Cartesian sign convention. For the biconvex lens \( R_1 = +30 \) cm and \( R_2 = -30 \) cm, and the object distance is u = −75 cm.
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
With \( \mu = 1.5 \):
\( \displaystyle \frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{30} + \frac{1}{30}\right) = \frac{1}{30}\,\mathrm{cm^{-1}} \)
\( \displaystyle \frac{1}{x} = \frac{1}{30} - \frac{1}{75} = \frac{5 - 2}{150} = \frac{1}{50} \)
\( x = 50\,\mathrm{cm} \)
With \( \mu = 1.6 \):
\( \displaystyle \frac{1}{f_2} = (1.6 - 1) \times \frac{2}{30} = \frac{1}{25}\,\mathrm{cm^{-1}} \)
\( \displaystyle \frac{1}{y} = \frac{1}{25} - \frac{1}{75} = \frac{3 - 1}{75} = \frac{2}{75} \)
\( y = 37.5\,\mathrm{cm} \)
\( x - y = 50\,\mathrm{cm} - 37.5\,\mathrm{cm} = 12.5\,\mathrm{cm} \)
A higher refractive index gives a shorter focal length, so the image forms closer to the lens.