Question

Radius of curvature of an equi – convex lens of glass (n = 1.5) is 30 cm. Find its focal length. An object of height 5.0 cm is placed at a distance of 60 cm from the optical centre of the lens. Find the position and the height of the image formed.

Answer 1

Given: n = 1.5, R₁ = + 30 cm, R₂ = - 30 cm, u = - 60cm

Using lens maker's formula,

`1/"f" = ("n - 1")[1/"R"_1 - 1/"R"_2]`

`1/"f" = (1.5 - 1)(1/30 + 1/30)`

`1/"f" = (0.5 xx 2)/30`

`1/"f" = 1/30`

∴ f = 30 cm

Linear magnification, 

`"m" = "f"/("f + u")`

`"m" = 30/(30 - 60)`

`"m" = 30/(- 30)`

m = - 1

and m = `v/u`

`- 1 = v/(- 60)`

v = 60

image will be formed at 60 cm from lens.

Also, m = `"height of image"/"height of object"`

or, - 1 = `"height of image"/5`

∴ height of image = - 5 cm.

Negative sign indicates that the image is inverted.