The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is at 1 m before the front surface of the lens. Calculate the distance of the image from the back surface of the lens.

Answer :

Answer:

-1.19 m

Explanation:

R1 = + 4 cm

R2 = - 15 cm

n = 1.5

distance of object, u = - 1 m

let the focal length of the lens is f and the distance of image is v.

use lens makers formula to find the focal length of the lens

[tex]\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]

By substituting the values, we get

[tex]\frac{1}{f}=\left ( 1.5-1 \right )\left ( \frac{1}{4}+\frac{1}{15} \right )[/tex]

[tex]\frac{1}{f}=\frac{19}{120}[/tex]   .... (1)

By using the lens equation

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{19}{120}=\frac{1}{v}+\frac{1}{1}[/tex]    from equation (1)

[tex]\frac{1}{v}=\frac{19-120}{120}[/tex]

v = -1.19 m

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