PART A: An object 1.15 cm high is placed 26.9 cm to the left of a converging len

Question

PART A:

An object 1.15 cm high is placed 26.9 cm to the left of a converging lens of focal length 14.2 cm. Find the position of the image of this object relative to this lens. Use a positive sign if the image lies to the right of the lens, and vice-versa.

PART B:

A second converging lens of focal length 14.3 cm is placed 11.6 cm to the right of the first lens. The image produced by the first lens now becomes the object for the second lens. Find this object distance, ie., the distance between the first image and lens 2. Use a positive sign if the position is to the left of lens 2, and vice-versa.

PART C:

How far from the second lens is the image that it forms? The value should be positive if the image lies to the right of lens 2, and vice-versa?

PART D:

What is the height of the final image? Use a positive sign if it upright, and vice-versa.

Explanation / Answer

f = + 14.2 cm

object distance, do = 26.9 cm

Using lens equation,

1/f = 1/do + 1/di

1/14.2 = 1/26.9 + 1/di

di = 30.1 cm .........Ans

B) object distance for second lens,

do' = 11.6 - 30.1 = - 18.5 cm

c) f = +14.3 cm

1/14.3 = 1/(-18.5) + 1/di'

di' = 8.07 cm

d) m1 = - (di / do) = - (30.1 / 26.9) = - 1.12

m2 = - (di' / do) = - (-8.07 / 18.5) = 0.436

magnification =m1 x m2 = - 0.49

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