Respuesta :
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
[tex]\frac{1}{p} + \frac{1}{q} = \frac{1}{f}[/tex]
[tex]\frac{1}{14} + \frac{1}{q} = \frac{1}{40}[/tex]
q = - 21.5 cm
using the equation for magnification as
[tex]\frac{i}{o} = \frac{-q}{p}[/tex]
[tex]\frac{i}{2} = \frac{-(- 21.5)}{14}[/tex]
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
[tex]\frac{1}{p} + \frac{1}{q} = \frac{1}{f}[/tex]
[tex]\frac{1}{41.5} + \frac{1}{q} = \frac{1}{20}[/tex]
q = 38.6 cm
image position : 38.6 cm
using the equation for magnification as
[tex]\frac{i}{o} = \frac{-q}{p}[/tex]
[tex]\frac{i}{3.1} = \frac{-(38.6)}{41.5}[/tex]
i = - 2.88 cm
image height = 2.8 cm
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
