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Physics Question on Ray optics and optical instruments

One plano-convex and one plano-concave lens of same radius of curvature ′R′'R' but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 11 is μ1\mu_1 and that of 22 is μ2\mu_2, then the focal length of the combination is :

A

R2−(μ1−μ2)\frac{R}{2-\left(\mu_{1} -\mu_{2}\right)}

B

2Rμ1−μ2\frac{ 2 R}{\mu_{1} -\mu_{2} }

C

R2(μ1−μ2)\frac{R}{2\left(\mu_{1} -\mu_{2}\right)}

D

Rμ1−μ2\frac{ R}{\mu_{1} -\mu_{2} }

Answer

Rμ1−μ2\frac{ R}{\mu_{1} -\mu_{2} }

Explanation

Solution

For 1st lens 1f1=(μ1−11)(1∞−1−R)=μ1−1R\frac{1}{f_{1}} = \left(\frac{\mu_{1} -1}{ 1}\right) \left(\frac{1}{\infty} - \frac{1}{-R}\right) =\frac{\mu_{1} -1}{ R} for 2nd lens 1f2=(μ2−11)(1−R−0)=−μ2−1R\frac{1}{f_{2}} = \left(\frac{\mu_{2} -1}{1}\right) \left(\frac{1}{-R} -0\right) =- \frac{\mu_{2} -1}{R} 1feq=1f1+1f2\frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} 1feq=Rμ1−1+R−(μ2−1)⇒1feq=Rμ1−μ2\frac{1}{f_{eq}} = \frac{R}{\mu_{1}-1} + \frac{R}{-\left(\mu_{2} -1\right)} \Rightarrow \frac{1}{f_{eq}} = \frac{R}{\mu_{1} -\mu_{2}} Hence feq=μ1−μ2Rf_{eq} = \frac{\mu_{1} -\mu_{2}}{R}