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  • Please help! A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following option best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following option best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

  • Option 1)

    Virtual, upright, height = 1 cm

  • Option 2)

    Virtual, upright, height = 0.5 cm

  • Option 3)

    Real, inverted, height = 4 cm

  • Option 4)

    Real, inverted, height = 1 cm

 

Answers (1)

best_answer

 

Thin lens formula -

frac{1}{v}-frac{1}{u}= frac{1}{f}

 

- wherein

u , and , v are object and image distance from lens.

 

 

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

 

 u=-30 cm

\frac{1}{f}=(\mu-1)\: (\frac{1}{R_{1}}-\frac{1}{R_{2}})

f=20 cm

\frac{1}{v}-\frac{1}{u}=\frac{1}{f} =>\frac{1}{V}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}

V=60 cm, real image 

m=\frac{V}{u}=\frac{h_{2}}{h_{1}} =>\frac{60}{-30}=\frac{h_{2}}{2cm} =>h_{2}=-4cm

\therefore Real Image, inverted, height =4 cm


Option 1)

Virtual, upright, height = 1 cm

This option is incorrect 

Option 2)

Virtual, upright, height = 0.5 cm

This option is incorrect 

Option 3)

Real, inverted, height = 4 cm

This option is correct 

Option 4)

Real, inverted, height = 1 cm

This option is incorrect 

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