Ray Optics

JEE Main level, single correct, medium

122 of 295

Question 122 of 295

A spherical interface lens of radius RR separates two media of refractive indices 11 and 1.41.4 respectively as shown in the figure below. A point source is placed at a distance of 4R4R in front of spherical interface. The magnitude of the magnification of point source image is _______.

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Answer: (A) 1.661.66

Given:
Refractive index of the first medium, n1=1n_1 = 1
Refractive index of the second medium, n2=1.4=75n_2 = 1.4 = \dfrac{7}{5}
Object distance, u=−4Ru = -4R
Radius of curvature of the spherical interface, R=+RR = +R (since the surface is convex towards the rarer medium)

Using the formula for refraction at a single spherical surface:
n2v−n1u=n2−n1R\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}

Substituting the given values:
1.4v−1−4R=1.4−1R\dfrac{1.4}{v} - \dfrac{1}{-4R} = \dfrac{1.4 - 1}{R}

75v+14R=0.4R\dfrac{7}{5v} + \dfrac{1}{4R} = \dfrac{0.4}{R}

75v+14R=25R\dfrac{7}{5v} + \dfrac{1}{4R} = \dfrac{2}{5R}

75v=25R−14R\dfrac{7}{5v} = \dfrac{2}{5R} - \dfrac{1}{4R}

75v=8−520R=320R\dfrac{7}{5v} = \dfrac{8 - 5}{20R} = \dfrac{3}{20R}

v=7×20R5×3=28R3v = \dfrac{7 \times 20R}{5 \times 3} = \dfrac{28R}{3}

The transverse magnification mm for a spherical refracting surface is given by:
m=n1vn2um = \dfrac{n_1 v}{n_2 u}

Substituting the values of n1n_1, n2n_2, uu, and vv:
m=1×(28R3)1.4×(−4R)m = \dfrac{1 \times \left(\dfrac{28R}{3}\right)}{1.4 \times (-4R)}

m=28R3−28R5m = \dfrac{\dfrac{28R}{3}}{-\dfrac{28R}{5}}

m=−53m = -\dfrac{5}{3}

The magnitude of the magnification is:
∣m∣=53≈1.66|m| = \dfrac{5}{3} \approx 1.66

Answer: 1.661.66

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