Question

The index of refraction of a thin lens is 1.5. Its one surface is convex (radius of curvature 20 cm) and the other planar. Calculate the focal length of the lens. What kind of an image does it form when the object is real and at 40 cm before the lens? What is the focal length if the lens is turned around? How does this influence image formation?

Answer 1

Answer:

f = 40 cm

image formed at a distance of 40 cm from lens is magnified and virtual.

when lens is turned around focal length is f = 40 cm

Explanation:

given data:

$R_1 = 20 cm$

$R_2 = \infty$

Refraction index of lens = 1.5

focal length of lens is given as

$\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})$

Putting all value to get focal length value

$\frac{1}{f} = (1.5 -1) \frac{1}{20}$

$\frac{1}{f} =\frac{0.5}{20}$

f = 40 cm

image formed at a distance of 40 cm from lens is magnified and virtual.

when lens is turned around

$R_1 = \infty$

$R_2 = -20cm$

Refraction index of lens = 1.5

focal length of lens is given as

$\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})$

Putting all value to get focal length value

$\frac{1}{f} = (1.5 -1) \frac{1}{20}$

$\frac{1}{f} =\frac{0.5}{20}$

f = 40 cm

there is no change can be seen between two condition. image will form at 40 cm from lens