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A lens having focal length f and aperture of diameter d forms an image of intensity I.

Apreture of diameter \frac{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:

  • Option 1)

    f and \frac{I}{4}

  • Option 2)

    \frac{3f}{4} and \frac{I}{2}

  • Option 3)

    f and \frac{3I}{4}

  • Option 4)

    \frac{f}{2}  and \frac{I}{2}

 

Answers (1)

best_answer

 

Lensmaker's Formula -

\frac{1}{f}= \left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}- \frac{1}{R_{2}}\right )
 

- wherein

\mu _{1}= refractive index of medium of object

\mu _{2}= refractive index of lens

R_{1}\, and \, R_{2} are radius of curvature of two surface

 

 Focal length of lens depends on material and its radius of curvature. Hence focal length will remain same

Area covered is \frac{1}{4} of total area hence \frac{1}{4} thof intensity will be blocked. 

Intensity received = \frac{3I}{4}

\therefore focal length =f

& intensity = \frac{3I}{4}


Option 1)

f and \frac{I}{4}

This is incorrect option 

Option 2)

\frac{3f}{4} and \frac{I}{2}

This is incorrect option 

Option 3)

f and \frac{3I}{4}

This is correct option 

Option 4)

\frac{f}{2}  and \frac{I}{2}

This is incorrect option 

Posted by

divya.saini

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