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NEET 2002
Exam
NEET
Year
2002
Subject
Physics
Chapter
Ray Optics

NEET 2002 · Physics · Ray Optics

A bulb is located on a wall. Its image is to be obtained on a...

A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance between parallel walls is ' dd ' then the required focal length of the lens placed in between the walls is:A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distanceA bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance

Options

  1. Aonly d4\frac{d}{4}
  2. Bonly d2\frac{d}{2}
  3. Cmore than d4\frac{d}{4} but less than d2\frac{d}{2}
  4. Dless than or equal to d4\frac{d}{4}

Correct answer

D. less than or equal to d4\frac{d}{4}

Step-by-step solution

A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance Since v+u=d⇒v=d−uv+u=d \Rightarrow v=d-u From lens formula 1f=1v−1u⇒1f=1d−u−1−u=d(d−u)u⇒f=du−u2d…(i)\begin{aligned}\frac{1}{f}& =\frac{1}{v}-\frac{1}{u}\\ \Rightarrow \quad \frac{1}{f}& =\frac{1}{d-u}-\frac{1}{-u}\\ & =\frac{d}{(d-u) u}\\ \Rightarrow \quad f & =\frac{d u-u^2}{d} \ldots(\mathrm{i}) \end{aligned}Differentiating eq. (i) we have dfdu=1d(\frac{d f}{d u}=\frac{1}{d}(d-2 u) for maximum value of f1=dfdu=0f_1=\frac{d f}{d u}=0⇒1d(d−2u)=0⇒d=24⇒u=d2\begin{aligned}& \Rightarrow \frac{1}{d}(d-2 u)=0 \Rightarrow d=24 \\ & \Rightarrow u=\frac{d}{2} \end{aligned}$puttingthesevalueineq.(i)f=putting these value in eq. (i) f=\frac{d}{4}$
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