Question

We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What must be the focal length of the appropriate positive lens? If it is a bi-convex lens of refractive index 1.5, what are the values of radii of curvature?

Answer 1

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

$\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$

$\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}$

f = 30 cm

using lens formula

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})$

$R_1 = R\ and\ R_2 = -R$

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})$

$R = (n -1)\ f$

$R = 2(1.5 -1)\ 30$

R = 30 cm

hence, the radii of curvature is 30 cm.