JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
A plano-convex lens (focal length \(f_2,\) refractive index \(\mu _2,\) radius of curvature \(R\) ) fits exactly into a plano-concave lens (focal length \(f_1,\) refractive index \(\mu _1,\) radius of curvature \(R\) ). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be
- A \(f_1\,-\,f_2\)
- B \(\frac{R}{{{\mu _2} - {\mu _1}}}\)
- C \(\frac{{2{f_1}{f_2}}}{{{f_1} + {f_2}}}\)
- D \({{f_1} + {f_2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{R}{{{\mu _2} - {\mu _1}}}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{f}=\frac{\left(\mu_{1}-1\right)}{R}+\frac{\left(1-\mu_{2}\right)}{R}\) \(\frac{1}{f}=\frac{\left(\mu_{1}-\mu_{2}\right)}{R}\) \(\Rightarrow \quad \mathrm{f}=\frac{\mathrm{R}}{\mu_{1}-\mu_{2}}\)
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