Question

An image four times the size of the object is formed on a screen located 10.0 cm from the object, by a thin lens. (a) Describe the type of lens is required. (b) Determine the focal length of the lens. () Where should the lens be positioned relative to the object? (d) If the lens is made of plastic with an index of refraction of 1.50, and one of its surfaces is flat, what is the radius of curvature of the other surface. Assume that the lens is placed in air. (e) Use array diagram to show how the image is formed by the lens.

Answer 1

Given:

The image formed is real of a real object.

M = magnification = -4 [magnification of real image of a real object is negative]

d = distance between object and image = 10cm

a) The lens required is a converging lens, as only a converging lens forms a real image of a real object.

b) let u = object distance

v = image distance

f = focal length

By convention, the real object distance is taken negatively, and the real image distance for the lens is taken positively.

Then, |v| + |u| = 10------(i)

also, magnification M = v/u = -4

=> v = -4u------(ii)

Therefore, using (i) and (ii), we get

|4u| + |u| = 10

=> |u| = 2 cm.

Therefore, u = -2 cm.

from (ii), v = -4*-2 = 8 cm

Using Lens equation:

f ' .

$\Rightarrow \frac{1}{f}=\frac{1}{8}-\frac{1}{-2}$

$=\frac{1}{8}+\frac{1}{2} = \frac{5}{8}$

=> f = 1.6 cm [answer]

c) The magnitude of object distance = 2 cm.

Therefore, the lens is positioned at a distance of 2 cm relative to the object. [answer]

d) n = refractive index of lens = 1.5

R = radius of curvature of one surface

the radius of curvature of another surface = infinity [since the radius of a flat surface is infinity]

Using Lens's maker equation:

1 = (n − 1) f R1 R2

$\Rightarrow \frac{1}{1.6}=(1.5-1)\left ( \frac{1}{R}-\frac{1}{\infty} \right )$

$\Rightarrow \frac{1}{1.6}=0.5*\frac{1}{R}$

$\Rightarrow R = 0.5*1.6 = 0.8 cm$ [answer]

d) see the diagram:

T F 1. Ge 2 cm 1. С. 8cm

O = object

F = focal point.

I = image