JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 24)
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.
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Answer
120
Explanation
$\frac{1}{f_{\text {concave }}}=(1.75-1)\left(\frac{1}{\infty}-\frac{1}{+30}\right)=-\frac{0.75}{30}$
$f_{\text {concave }}=-40 \mathrm{~cm}$
$\frac{1}{f_{\text {convex }}}=(1.75-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.75}{30}$
$f_{\text {convex }}=40 \mathrm{~cm}$
Let the first image is formed at $v_{1}$ so
$\frac{1}{v_{1}}-\frac{1}{\infty}=\frac{1}{f_{\text {concave }}}=-\frac{1}{40}$
$\Rightarrow v_{1}=-40 \mathrm{~cm}$
for second image
$\frac{1}{x-40}-\frac{1}{-80}=\frac{1}{40}$
$\Rightarrow x=120 \mathrm{~cm}$
$f_{\text {concave }}=-40 \mathrm{~cm}$
$\frac{1}{f_{\text {convex }}}=(1.75-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.75}{30}$
$f_{\text {convex }}=40 \mathrm{~cm}$
Let the first image is formed at $v_{1}$ so
$\frac{1}{v_{1}}-\frac{1}{\infty}=\frac{1}{f_{\text {concave }}}=-\frac{1}{40}$
$\Rightarrow v_{1}=-40 \mathrm{~cm}$
for second image
$\frac{1}{x-40}-\frac{1}{-80}=\frac{1}{40}$
$\Rightarrow x=120 \mathrm{~cm}$
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