Q133 / 200 · NEET 2010 (Screening) · Physics

A lens having focal length $f$ and aperture of diameter $\mathrm{d}$ forms an image of intensity $I$. Aperture of diameter $\frac{\mathrm{d}}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

  • A
    $f$ and $\frac{1}{4}$
  • B
    $\frac{3 f}{4}$ and $\frac{I}{2}$
  • Correct answer
    C
    f and $\frac{3 I}{4}$
  • D
    $\frac{\mathrm{f}}{2}$ and $\frac{\mathrm{I}}{2}$

Solution

Intensity, $I \propto A^2$ $\Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=\left[\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right]^2=\frac{\pi \mathbf{r}^2-\frac{\pi \mathbf{r}^2}{4}}{\pi \mathbf{r}^2}=\frac{3}{4}$ $\Rightarrow \quad \mathrm{I}_2=\frac{3}{4} \mathrm{I}_1$ and focal length remains unchanged.

Hence, the correct answer is option: 'C'.

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