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Question: Parallel rays from medium of refractive index 3µ are incident on lens and converge at distance *f* f...

Parallel rays from medium of refractive index 3µ are incident on lens and converge at distance f from lens in medium of refractive index 2µ. Then f is

A

−4R5\frac{-4R}{5}

B

2R

C

3R2\frac{3R}{2}

D

−R2-\frac{R}{2}

Answer

−4R5\frac{-4R}{5}

Explanation

Solution

The lens maker's formula for a lens in a medium is given by: n3f=n2−n1R1+n3−n2R2\frac{n_3}{f} = \frac{n_2 - n_1}{R_1} + \frac{n_3 - n_2}{R_2} Given: n1=3μn_1 = 3\mu, n2=μn_2 = \mu, n3=2μn_3 = 2\mu, R1=RR_1 = R, R2=−2RR_2 = -2R. Substituting these values: 2μf=μ−3μR+2μ−μ−2R\frac{2\mu}{f} = \frac{\mu - 3\mu}{R} + \frac{2\mu - \mu}{-2R} 2μf=−2μR+μ−2R\frac{2\mu}{f} = \frac{-2\mu}{R} + \frac{\mu}{-2R} 2μf=−2μR−μ2R=−4μ+μ2R=−5μ2R\frac{2\mu}{f} = -\frac{2\mu}{R} - \frac{\mu}{2R} = -\frac{4\mu + \mu}{2R} = -\frac{5\mu}{2R} 2f=−52R\frac{2}{f} = -\frac{5}{2R} f=4R−5=−4R5f = \frac{4R}{-5} = -\frac{4R}{5}