JEE MAIN - Physics (2025 - 8th April Evening Shift - No. 2)
A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ________.
$ \frac{60}{7} $
15
45
30
Explanation
Equivalent focal length
$$\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\ & =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\ & \mathrm{f}=-60 \mathrm{~cm} \end{aligned}$$
Lens formula
$$\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\ & \mathrm{v}=-15 \mathrm{~cm} \end{aligned}$$
Comments (0)
