Question

An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of 40 cm and 60 cm. The glass has an index of refraction of 1.54. What is the focal length of this lens in air?

Answer 1

Answer:

44.4cm

Explanation:

glass has an index of refraction .n = 1.54

radii of curvature of 40 cm R1 = 40 by

radii of curvature of 600 cm R2 = 60

Now, by lens maker formula

1/f = (n - 1) (1/R1 - 1/R2)

Putting in the given values for n = 1.54 , we get f = 22.2

$\frac{1}{f} = (1.54 -1) (\frac{1}{40cm} -\frac{1}{(-60cm)} )$

$\frac{1}{f} = 0.0225$

f = 1 / 0.0225

f = 44.4cm

so, focal length in air will be  = 44.4 cm