JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 27)
A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses $$\mathrm{L}_{1}$$ and $$\mathrm{L}_{2}$$ with focal length $$24 \mathrm{~cm}$$ and $$9 \mathrm{~cm}$$ respectively. The distance between two lenses is $$10 \mathrm{~cm}$$ and the object is placed $$6 \mathrm{~cm}$$ away from lens $$\mathrm{L}_{1}$$ as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ $$\mathrm{cm}$$.
_10th_April_Evening_Shift_en_27_1.png)
Answer
34
Explanation
From Ist lens :
$\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}$
$$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\\\ \mathrm{v} & =-8 \mathrm{~cm} \end{aligned} $$
From IInd lens :
$\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}$
$$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{18} \\\\ \mathrm{v} & =18 \end{aligned} $$
_10th_April_Evening_Shift_en_27_2.png)
So distance between object and its image
= 6 + 10 + 18 = 34 cm
$\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}$
$$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\\\ \mathrm{v} & =-8 \mathrm{~cm} \end{aligned} $$
From IInd lens :
$\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}$
$$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{18} \\\\ \mathrm{v} & =18 \end{aligned} $$
So distance between object and its image
= 6 + 10 + 18 = 34 cm
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