Question

Sumati did some preliminary experiment with the lenses and found out that the
magnification of the eyepiece (L2) is 3. If in her experiment with L2 she found
an image at 24 cm from the lens, at what distance did she put the object?

Answer 1

Explanation:

m=v/u

here magnification(m)=3 and image distance (v)=24cm

therefore:

3=24/u

3u=24

u=8cm

At a distance of 8cm the object should be placed

Answer 2

Given: Sumati did some preliminary experiment with the lenses and found out that the magnification of the eyepiece (L2) is 3. In her experiment with L2 she found an image at 24 cm from the lens.

To find: Distance at which she put the object

Solution: Let the magnification of the lens be m, object distance be u and image distance by v.

Magnification of the lens tells about the height of the image as compared to the height of the object.

When, m>1 which means the image is larger than the object and when m<1, the image is smaller than the object.

m= 3

v= 24

The formula for magnification in a lens is given by the formula:

m = v/u

=> 3 = 24/u

=> u = 24/3

=> u = 8 cm

Therefore, Sumati had put the object 8 cm in front of the lens.