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Physics Question on Ray optics and optical instruments

The effective focal length of the lens combination shown in the figure in - 60 cm. The radii of curvature of the curved surfaces of the Plano-convex lenses are 12 cm each and refractive index of the material of the lens is 1.5. The refractive index of the liquid is:

A

1.33

B

1.42

C

1.53

D

1.6

Answer

1.6

Explanation

Solution

The system is equivalent to combination of three lenses in contact, i.e., 1F=1f1+1f2+1f3\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}} But by lens Makers formula 1f1=(1.5−1)(1−∞−1−12)=0.5×112=124\frac{1}{{{f}_{1}}}=(1.5-1)\left( \frac{1}{-\infty }-\frac{1}{-12} \right)=0.5\times \frac{1}{12}=\frac{1}{24} 1f2=(μ−1)(1−12−112)=−(μ−16)\frac{1}{{{f}_{2}}}=(\mu -1)\left( \frac{1}{-12}-\frac{1}{12} \right)=-\left( \frac{\mu -1}{6} \right) 1f3=(1.5−1)(112−1∞)=124\frac{1}{{{f}_{3}}}=(1.5-1)\left( \frac{1}{12}-\frac{1}{\infty } \right)=\frac{1}{24} Also F=−60 cmF=-60\,cm ∴\therefore 1−60=124−(μ−1)6+124=−(μ−1)6+112\frac{1}{-60}=\frac{1}{24}-\frac{(\mu -1)}{6}+\frac{1}{24}=-\frac{(\mu -1)}{6}+\frac{1}{12} or (μ−1)6=112+160\frac{(\mu -1)}{6}=\frac{1}{12}+\frac{1}{60} or (μ−1)6=5+160+660\frac{(\mu -1)}{6}=\frac{5+1}{60}+\frac{6}{60} or (μ−1)=3660=0.6(\mu -1)=\frac{36}{60}=0.6 or μ=0.6+1=1.6\mu =0.6+1=1.6