Question

The radii of curvature of a convex lens are 40cm and 10cm for each surface respectively. If the refractive index is 3/2 then the focal length of the lens is

Answer 1

Answer:

plzz give me brainliest ans and plzzzz follow me

Answer 2

Explanation:

We know that, focal depends upon the radius of curvature and refractive index.

here :

radius of curvature (R1) = 40 cm

Radius of curvature (R2) = - 10 cm.

refractive index (n) = 3/2

so,

$\frac{1}{f} = (n - 1)(\frac{1}{r1 } - \frac{1}{r2)} \\ \frac{1}{f} = (\frac{3}{2} - 1)( \frac{1}{40} - \frac{1}{</u></em></strong><strong><em><u>-</u></em></strong><strong><em><u>10} ) \\ \frac{1}{f} = ( \frac{3 - 2 }{2} )( \frac{1</u></em></strong><strong><em><u>+</u></em></strong><strong><em><u>4}{40} ) \\ \frac{1}{f} = \frac{1}{2} \times ( \frac{ </u></em></strong><strong><em><u>5</u></em></strong><strong><em><u>}{40} ) \\ \frac{1}{f} = \frac{ </u></em></strong><strong><em><u>1</u></em></strong><strong><em><u>}{</u></em></strong><strong><em><u>16</u></em></strong><strong><em><u>} \\ f = \frac{ </u></em></strong><strong><em><u>16</u></em></strong><strong><em><u>}{</u></em></strong><strong><em><u>1</u></em></strong><strong><em><u>}$