JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2023
- Hard
- Numerical
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is \(30\,cm\) and refraction index of the material for both the lenses is \(1.75\). Both the lenses are placed at distance of \(40\,cm\) from each other. Due to the combination, the image of the object is formed at distance \(x =..........cm\), from concave lens.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(120\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{ f _1}=(1.75-1)\left(-\frac{1}{30}\right)\)
\(\Rightarrow f _1=-40\,cm\)
\(\frac{1}{ f _2}=(1.75-1)\left(\frac{1}{30}\right) \Rightarrow f _2=40\,cm\)
Image from \(L_1\) will be virtual and on the left of \(L_1\) at focal length \(40 \,cm\). So the object for \(L_2\) will be \(80\,cm\) from \(L _2\) which is \(2 f\). Final image is formed at \(80\,cm\) from \(L _2\) on the right. So \(x=120\)
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