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To find the focal length of a convex mirror, a student records the following data :

Object Pin

 Convex Lens Convex Mirror

Image Pin
22.2 cm 32.2 cm 45.8 cm 71.2 cm

The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small,  f1 and f2 are close to :

  • Option 1)

    f_{1} = 12.7\, cm\;\; \; \; \; \; \; f_{2}=7.8\, cm

  • Option 2)

    f_{1} = 7.8\, cm\;\; \; \; \; \; \; f_{2}=12.7\, cm

  • Option 3)

    f_{1} = 7.8\, cm\;\; \; \; \; \; \; f_{2}=25.4\, cm

  • Option 4)

    f_{1} = 15.6\, cm\;\; \; \; \; \; \; f_{2}=25.4\, cm

 

Answers (1)

As we learnt in

Thin lens formula -

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

 

- wherein

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

 

 Taking f2 = 12.07

USe mirror formula

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{12.7}=\frac{1}{25.4}+\frac{1}{u}=\frac{1}{12.7}-\frac{1}{25.4}=\frac{1}{u}

u = 25.4 = v'

Now using len's formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}=\frac{1}{24.4+13.6}+\frac{1}{10}

\frac{1}{f_{1}}=\frac{1}{39}+\frac{1}{10}\ \; \Rightarrow\ \;f_{1}=\frac{390}{49}=7.96

Correct option is 2.

 


Option 1)

f_{1} = 12.7\, cm\;\; \; \; \; \; \; f_{2}=7.8\, cm

This is an incorrect option.

Option 2)

f_{1} = 7.8\, cm\;\; \; \; \; \; \; f_{2}=12.7\, cm

This is the correct option.

Option 3)

f_{1} = 7.8\, cm\;\; \; \; \; \; \; f_{2}=25.4\, cm

This is an incorrect option.

Option 4)

f_{1} = 15.6\, cm\;\; \; \; \; \; \; f_{2}=25.4\, cm

This is an incorrect option.

Posted by

Vakul

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