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QJEEM23Ph2791: JEE Main 2023 Physics - Class 12 JEEM23Ph2791 Solution: As shown in the figure, a combination of a thin plano concave…

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.

JEE Main 2023 (Online) 24th January Morning Shift Physics - Geometrical Optics Question 63 English

Step-by-Step Verified Solution

1fconcave =(1.75−1)(1∞−1+30)=−0.7530\frac{1}{f_{\text {concave }}}=(1.75-1)\left(\frac{1}{\infty}-\frac{1}{+30}\right)=-\frac{0.75}{30}

fconcave =−40 cmf_{\text {concave }}=-40 \mathrm{~cm}

1fconvex =(1.75−1)(130−1∞)=0.7530\frac{1}{f_{\text {convex }}}=(1.75-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.75}{30}

fconvex =40 cmf_{\text {convex }}=40 \mathrm{~cm}

Let the first image is formed at v1v_{1} so

1v1−1∞=1fconcave =−140\frac{1}{v_{1}}-\frac{1}{\infty}=\frac{1}{f_{\text {concave }}}=-\frac{1}{40}

⇒v1=−40 cm\Rightarrow v_{1}=-40 \mathrm{~cm}

for second image

1x−40−1−80=140\frac{1}{x-40}-\frac{1}{-80}=\frac{1}{40}

⇒x=120 cm\Rightarrow x=120 \mathrm{~cm}

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