Three lenses of focal lengths +10 cm, -10 c | vedclass

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(B) From the diagram,the object $O$ is at $30 \ cm$ from lens $A$. Lens $B$ is at $5 \ cm$ from lens $A$,and lens $C$ is at $10 \ cm$ from lens $B$.Fo

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) From the diagram,the object $O$ is at $30 \ cm$ from lens $A$. Lens $B$ is at $5 \ cm$ from lens $A$,and lens $C$ is at $10 \ cm$ from lens $B$.For lens $A$: $u_A = -30 \ cm$,$f_A = +10 \ cm$.Using the lens formula $\frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f_A}$:$\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{10} \Rightarrow \frac{1}{v_A} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \Rightarrow v_A = 15 \ cm$.The image formed by lens $A$ acts as an object for lens $B$. The distance of this image from lens $B$ is $u_B = +(v_A - 5) = +(15 - 5) = +10 \ cm$.For lens $B$: $u_B = +10 \ cm$,$f_B = -10 \ cm$.Using the lens formula: $\frac{1}{v_B} - \frac{1}{10} = \frac{1}{-10} \Rightarrow \frac{1}{v_B} = 0 \Rightarrow v_B = \infty$.The rays emerging from lens $B$ are parallel,so they act as an object at infinity for lens $C$.For lens $C$: $u_C = \infty$,$f_C = +30 \ cm$.Using the lens formula: $\frac{1}{v_C} - \frac{1}{\infty} = \frac{1}{30} \Rightarrow v_C = 30 \ cm$.The final image is formed at $30 \ cm$ from lens $C$. The total distance of lens $C$ from the object is $30 + 5 + 10 = 45 \ cm$. The final image is at $30 \ cm$ from lens $C$,so the distance from the object is $45 + 30 = 75 \ cm$.

Three lenses of focal lengths $+10 \ cm, -10 \ cm$ and $+30 \ cm$ are placed at distances of $30 \ cm, 35 \ cm$ and $45 \ cm$ respectively from an object. The distance between the object and the final image formed is: (in $cm$)

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