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A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of convex lens. If the distance between parallel walls is 'd' then required focal length of lens placed in between the walls is : -
A
Only $\dfrac{d}{4}$
B
Only $\dfrac{d}{2}$
C
More than $\dfrac{d}{4}$ but less than $\dfrac{d}{2}$
D
Less than or equal to $\dfrac{d}{4}$
Detailed Solution
For a real image of the bulb on the other wall, the object distance u and image distance v (magnitudes) must add up to d: $u + v = d$.
Lens formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \Rightarrow f = \dfrac{uv}{u + v} = \dfrac{uv}{d}$
For fixed $u + v = d$, the product uv is greatest when $u = v = \dfrac{d}{2}$, giving $uv_{max} = \dfrac{d^2}{4}$.
So $f = \dfrac{uv}{d} \leq \dfrac{d^2/4}{d} = \dfrac{d}{4}$
Equivalently, the minimum distance between an object and its real image is 4f, so $d \geq 4f$.
So the focal length must be less than or equal to $\dfrac{d}{4}$.
Lens formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \Rightarrow f = \dfrac{uv}{u + v} = \dfrac{uv}{d}$
For fixed $u + v = d$, the product uv is greatest when $u = v = \dfrac{d}{2}$, giving $uv_{max} = \dfrac{d^2}{4}$.
So $f = \dfrac{uv}{d} \leq \dfrac{d^2/4}{d} = \dfrac{d}{4}$
Equivalently, the minimum distance between an object and its real image is 4f, so $d \geq 4f$.
So the focal length must be less than or equal to $\dfrac{d}{4}$.