An object .69 cm high is placed 23.1 cm to the left of a converging lens of foca

Question

An object .69 cm high is placed 23.1 cm to the left of a converging lens of focal length 11.2 cm. Find the position of the image of this object relative to this lens.

A second converging lens of focal length 10.3 cm is placed 12.6 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, i.e., the distance between the first image and the lens 2.

How far from the second lens is the image that it forms?

What is the height of the final image?

Explanation / Answer

for converging lens,

f = +11.2 cm

object distance, do = 23.1 cm

Using lens equation,

1/f = 1/do + 1/di

1/11.2 = 1/23.1 + 1/di

di = 21.74 cm .............Ans

-----------------

now this image will work as object for second lens.

so object distance, do2 = 12.6 - 21.74 = - 9.14cm (object from second lens)

f = +10.3 cm

1/10.3 = 1/(-9.14) +1/di2

di2 = 4.84 cm ...........Ans (Image from second lens )

m = m1 m2 = (di2/do2) (di/do) = (4.84 / -9.14 ) ( 21.74 / 23.1) = - 0.5

m = hi / ho

- 0.5 = hi / 0.69

hi = - 0.34 cm (minus sign indicates inverted) ..........Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.