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QJEEM23Ph2472: JEE Main 2023 Physics - Class 12 JEEM23Ph2472 Solution: A point object, 'O' is placed in front of two thin symmetrical…

A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses L1\mathrm{L}_{1} and L2\mathrm{L}_{2} with focal length 24 cm24 \mathrm{~cm} and 9 cm9 \mathrm{~cm} respectively. The distance between two lenses is 10 cm10 \mathrm{~cm} and the object is placed 6 cm6 \mathrm{~cm} away from lens L1\mathrm{L}_{1} as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ cm\mathrm{cm} .

JEE Main 2023 (Online) 10th April Evening Shift Physics - Geometrical Optics Question 52 English

Step-by-Step Verified Solution

From Ist lens :

1v+16=124\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}

1vamp;=124−16=−18vamp;=−8 cm\begin{aligned}\frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\\\\mathrm{v} & =-8 \mathrm{~cm}\end{aligned}

From IInd lens :

1v+118=19\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}

1vamp;=118vamp;=18\begin{aligned}\frac{1}{\mathrm{v}} & =\frac{1}{18} \\\\\mathrm{v} & =18\end{aligned}

JEE Main 2023 (Online) 10th April Evening Shift Physics - Geometrical Optics Question 52 English Explanation

So distance between object and its image

= 6 + 10 + 18 = 34 cm

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