CBSE 2025, Compartment, Set 1

Question

A convex lens is kept coaxially on the left side of a concave mirror at a distance of $5f$ from it where $f$ is focal length of each of them. An object is kept at a distance of $2f$ on the left side of the convex lens. Find the distance of the final image from the mirror.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** **Step 1: Refraction through the convex lens** * Object distance, $u = -2f$ * Focal length, $f_{lens} = +f$ Using the lens formula $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$: $$\frac{1}{v_1} - \frac{1}{-2f} = \frac{1}{f}$$ $$\frac{1}{v_1} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}$$ $$v_1 = +2f$$ The first image ($I_1$) is formed at a distance $2f$ to the right of the convex lens. **Step 2: Reflection from the concave mirror** The total distance between the lens and mirror is $5f$. Thus, the image $I_1$ acts as a real object for the concave mirror. Distance of $I_1$ from the mirror, $u_m = -(5f - 2f) = -3f$. Focal length of the concave mirror, $f_m = -f$. Using the mirror formula $\frac{1}{v_2} + \frac{1}{u_m} = \frac{1}{f_m}$: $$\frac{1}{v_2} + \frac{1}{-3f} = \frac{1}{-f}$$ $$\frac{1}{v_2} = -\frac{1}{f} + \frac{1}{3f}$$ $$\frac{1}{v_2} = \frac{-3 + 1}{3f} = -\frac{2}{3f}$$ $$v_2 = -\frac{3}{2}f = -1.5f$$ The final image is formed at a distance of **$1.5f$ in front of the concave mirror**.

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Physics > Optics > Ray Optics And Optical Instruments

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