A biconvex lens of focal length 15 ,cm is in | vedclass

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(C) Step $1$: Refraction through the lens.Using the lens formula: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30 \,cm$ and $f = +15 \,cm$.

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virtual and at a distance of $20 \,cm$ from the mirror

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(C) Step $1$: Refraction through the lens.Using the lens formula: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30 \,cm$ and $f = +15 \,cm$.$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{15} \implies \frac{1}{v_1} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$.So,$v_1 = +30 \,cm$. This image acts as a virtual object for the mirror.Step $2$: Reflection from the mirror.The distance of this image from the lens is $30 \,cm$ to the right. Since the mirror is $10 \,cm$ from the lens,the image is $30 - 10 = 20 \,cm$ behind the mirror.For a plane mirror,the object distance $u_m = +20 \,cm$ (virtual object).The image formed by the plane mirror will be at $v_m = -20 \,cm$ (real image in front of the mirror).Step $3$: Refraction again through the lens.The light rays reflected from the mirror travel back through the lens. The object for the lens is now at $10 + 20 = 30 \,cm$ to the left of the lens $(u_2 = -30 \,cm)$.Using the lens formula again: $\frac{1}{v_2} - \frac{1}{-30} = \frac{1}{15} \implies v_2 = +30 \,cm$.This image is $30 \,cm$ to the right of the lens,which is $30 - 10 = 20 \,cm$ behind the mirror. Since the light rays are converging towards a point behind the mirror,the final image is virtual and located $20 \,cm$ from the mirror.

$A$ biconvex lens of focal length $15 \,cm$ is in front of a plane mirror. The distance between the lens and the mirror is $10 \,cm$. $A$ small object is kept at a distance of $30 \,cm$ from the lens. The final image is

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