Question

Two converging lenses with focal lengths of 40 cm and 20 cm are 16 cm apart. A 2.0 cm -tall object is 14 cm in front of the 40 cm -focal-length lens.

Calculate the image position.

Calculate the image height.

Answer 1

Answer:

38.6 cm

2.8 cm

Explanation:

For converging lens of 40 cm :

f = focal length = 40 cm

p = object distance = 14 cm

q = image distance

o = height of the object = 2 cm

i = image height

using the lens equation

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$

$\frac{1}{14} + \frac{1}{q} = \frac{1}{40}$

q = - 21.5 cm

using the equation for magnification as

$\frac{i}{o} = \frac{-q}{p}$

$\frac{i}{2} = \frac{-(- 21.5)}{14}$

i = 3.1 cm

For converging lens of 20 cm :

f = focal length = 20 cm

p = object distance = 21.5 + 20 = 41.5 cm

q = image distance

o = height of the object = 3.1 cm

i = image height

using the lens equation

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$

$\frac{1}{41.5} + \frac{1}{q} = \frac{1}{20}$

q = 38.6  cm

image position : 38.6 cm

using the equation for magnification as

$\frac{i}{o} = \frac{-q}{p}$

$\frac{i}{3.1} = \frac{-(38.6)}{41.5}$

i = - 2.88 cm

image height = 2.8 cm

Answer 2

The position of the image is 38.6 cm and the image height is 2.88 cm.

Image position

The position of the image is determined by using lens formula as shown below;

1/f = 1/q + 1/p

1/q = 1/f - 1/p

1/q = 1/40 - 1/14

1/q = -0.0464

q = -21.54 cm

i/2 = -q/p

i/2 = -(-21.54)/14

i = 3.1 cm

New object distance due to second lens, = 20 cm + 21.54 cm = 41.54 cm

1/q = 1/f - 1/p

1/q = 1/20 - 1/41.54

q = 38. 6 cm

Image height

The height of the image is calculated as follows;

i/3.1 = -(-38.6)/41.54

i = 2.88 cm

Learn more about lens formula here: https://brainly.com/question/25876096