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(B) Let the distance between the object (bulb) and the screen (opposite wall) be $D = 3\, m$. Let the distance of the lens from the object be $u$ and

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(B) Let the distance between the object (bulb) and the screen (opposite wall) be $D = 3\, m$. Let the distance of the lens from the object be $u$ and from the screen be $v$. Then $u + v = D = 3\, m$. From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Using sign convention,$u$ is negative,so $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{u+v}{uv}$. Substituting $u+v = D$,we get $f = \frac{uv}{D}$. Since $v = D - u$,we have $f = \frac{u(D-u)}{D} = \frac{Du - u^2}{D}$. For a real image to be formed on the screen,the condition $D \geq 4f$ must be satisfied. Therefore,$f \leq \frac{D}{4}$. Substituting $D = 3\, m$,we get $f_{\max} = \frac{3}{4} = 0.75\, m$.

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $3\, m$ away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

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