Question

The radii of curvature of a convex lenses are 0.05m and 05.05m. Refractive index is 1.5 . Its focal length is____

Answer 1

Hope u like my process
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The len's maker formula to be used is

$= > \bf \frac{1}{f} = ( \mu - 1)( \frac{1}{ r_{1} } - \frac{1}{r_{2} } )$
Where,

=> f = focal length;

=> R1 and R2 are radius of curvatures

$= > \bf \mu = refractive \: \: index$
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Given:-
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$= > \bf \: r_{1} = 0.05 \: \: m \\ \\ = > \bf r_{2} = 05.05 \: \: m \\ \\ = > \bf \mu = 1.5$

Thus,

Now,
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$\bf = > \frac{1}{f} = ( \mu - 1)( \frac{1}{ r_{1} } - \frac{1}{ r_{2} } ) \\ \\ = > \it\frac{1}{f} = (1.5 - 1)( \frac{1}{0.05} - \frac{1}{5.05} ) \\ \\ = > \it \frac{1}{f} = \frac{0.5}{5} ( \frac{100}{1} - \frac{100}{101} ) \\ \\ = > \it \: \frac{1}{f} = 100 \times 0.1(1 - \frac{1}{101} ) \\ \\ = > \it \frac{1}{f} = 10( \frac{101 - 1}{101} ) = \frac{10 \times 100}{101} \\ \\ = > \it f = \frac{101}{1000} = \bf \underline{ \: 0.101 \: \: m \: }$
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So the required focal length is

=> 0.101 m, or, 101 cm

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Hope this is ur required answer

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Answer 2

Given:

refractive index μ = 1.5

radius of curvatures r₁ = 0.05m

                                  r₂=5.05m

To Find :

Find the focal length , f

Solution:

Focal length:

The focal length of a convex lens is the distance between the center of a lens and its focus. The focal length of an optical instrument/object is a measure of how strongly/sharply the system converges/diverges light and it is just the inverse of the optical power of the system.

we know that focal length of the convex lens is given by,

1/f = (μ -1) {(1/r₁) - (1/r₂)}

putting the values of μ , r₁ , r₂

1/f = (1.5 - 1) { ( 1/0.05)  -  ( 1/ 5.05) }

1/f = (0.5) { 20 - 0.198}

1/f = (0.5) {19.80}

1/f = 9.901

f = 0.101

Hence the focal length of the convex lens is 0.101 m