A biconvex lens has a radius of curvature of magnitude `20cm`. Which one of the following options describes best the image formed of an object of height `2cm` place `30 cm` from the lens ?

Options

A. Virtual , upright, height `= 1 cm`
B. Virtual , upright, height `= 0.5 cm`
C. Real, inverted, height `= 4 cm`
D. Real, inverted, height `= 1 cm`

Answer

C

Explanation

Here, `R_(1) = + 20 cm , R_(2) = - 20 cm`
[Note. In this question, the value of refractive index of lens is not given. Let `mu = 3//2`]
Using lens maker's formula
`(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`= ((3)/(2) - 1)((1)/(20) - (1)/(-20)) = (1)/(20)`
`f = 20 cm`
Here, `u = - 30 cm, f = 20 cm, v = ?`
Using lens formula, we have
`(1)/(f) = (1)/(v) - (1)/(u)` or `(1)/(20) = (1)/(v) - (1)/(-30) = (1)/(v) + (1)/(30)`
or `(1)/(v) = (1)/(20) - (1)/(30) = (1)/(60)` or `v = 60 cm`
Magnification, `m = (v)/(u) = (h_(i))/(h_(0))`
`(60 cm)/(-30cm) = (h_(i))/(2 cm)` or `h_(i) = - 4cm`
it means. the image is real, inverted and of height `4 cm`. The path of rays is shown in Fig.