A converging lens with a focal length of 10 cm is held over an empty bucket so t

ID: 1263861 • Letter: A

Question

A converging lens with a focal length of 10 cm is held over an empty bucket so that it images a rock at the bottom of the bucket.

(a) If the lens is 25 cm above the bottom of the bucket, where is the image located? What is the magnification? Is it real or virtual?

(b) To what depth must we add water to the bucket if we want to make the image larger than the object? The lens stays dry in this procedure.

Hi, can someone assist me with the second part? I already figured out the first part. I got:

di = 16.7cm

m= -0.668 The image is real

But I don't understand the second question. Thanks in advance

Given that-

f=10cm, u=25cm

1/v+1/u=1/f

1/v=1/10-1/25

image located at v=16.67cm

magnification m=-v/u=-0.667 (real)

We know that magnification m=-v/u=I/O

As we add the water to the bucket it increase the level of rock.

we know that Apparent depth = Actual depth/refractive index.

Apparent depth=25/1.33=18.79cm

This Apparent depth=new distance of object u'=18.79.

Now 1/v=1/f-1/u=1/10-1/18.79

v=21.37

now m=-v/u=1.13=I/O

I=1.13O here image is larger.

Hence we should add water at height h=u-u'=25-18.79=6.21cm

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