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(A) $1$. For the first lens $(f_1 = +10 \, cm)$: The object distance $u_1 = -30 \, cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) $1$. For the first lens $(f_1 = +10 \, cm)$: The object distance $u_1 = -30 \, cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,we get $\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \Rightarrow \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$. Thus,$v_1 = +15 \, cm$.$2$. The image formed by the first lens acts as an object for the second lens $(f_2 = -10 \, cm)$. The distance between the first and second lens is $5 \, cm$. So,the object distance for the second lens is $u_2 = +(15 - 5) = +10 \, cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$,we get $\frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10} \Rightarrow \frac{1}{v_2} = 0$,which means $v_2 = \infty$.$3$. The parallel rays from the second lens fall on the third lens $(f_3 = +30 \, cm)$. Since the rays are parallel,the image is formed at the focus of the third lens. The distance between the second and third lens is $10 \, cm$. The image is formed at $30 \, cm$ from the third lens.$4$. The total distance from the object $O$ is $30 \, cm$ (to first lens) $+ 5 \, cm$ (between first and second) $+ 10 \, cm$ (between second and third) $+ 30 \, cm$ (from third lens) $= 75 \, cm$.

Find the distance of the final image from the object $O$,formed by the combination of lenses shown in the figure (in $cm$):

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