Ray Optics

JEE Main level, single correct, medium

127 of 295

Question 127 of 295

Refer the figure given below. μ1\mu_1 and μ2\mu_2 are refractive indices of air and lens material. The height of image will be _______ cm.

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Answer: (A) 11

Taking the pole PP as the origin and the direction of incident light as positive:
μ1=1\mu_1 = 1 (air), μ2=1.5\mu_2 = 1.5 (lens)
u=−40 cmu = -40\text{ cm}
R=−20 cmR = -20\text{ cm} (centre of curvature lies to the left of PP)
ho=+2 cmh_o = +2\text{ cm}

Using the formula for refraction at a single spherical surface:
μ2v−μ1u=μ2−μ1R\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}

1.5v−1−40=1.5−1−20\dfrac{1.5}{v} - \dfrac{1}{-40} = \dfrac{1.5 - 1}{-20}

1.5v+140=−140\dfrac{1.5}{v} + \dfrac{1}{40} = -\dfrac{1}{40}

1.5v=−240=−120\dfrac{1.5}{v} = -\dfrac{2}{40} = -\dfrac{1}{20}

v=−30 cmv = -30\text{ cm}

The negative sign indicates a virtual image formed 30 cm30\text{ cm} to the left of PP.

The transverse magnification for refraction at a spherical surface is:
m=μ1⋅vμ2⋅u=1⋅(−30)1.5⋅(−40)=−30−60=0.5m = \dfrac{\mu_1 \cdot v}{\mu_2 \cdot u} = \dfrac{1 \cdot (-30)}{1.5 \cdot (-40)} = \dfrac{-30}{-60} = 0.5

Therefore:
hi=m⋅ho=0.5×2=1 cmh_i = m \cdot h_o = 0.5 \times 2 = 1\text{ cm}

Hence, the correct option is (1) 1(1)\ 1.

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