In an experiment ,an object was placed on the principal axis of a convex lens 25 centimetres away from the lens. Areal image 4 times the size of the object was obtained. The focal length of the lens is?

Answer 1

We can use the magnification formula:

m = -v/u

where m is the magnification, v is the image distance, and u is the object distance.

We are given that the image is 4 times the size of the object, so m = 4. We are also given that the object is 25 cm away from the lens, so u = -25 cm (since the object is in front of the lens, the distance is negative). Finally, we want to find the focal length f of the lens.

Using the lens formula:

1/f = 1/v - 1/u

We can rearrange to solve for v:

1/v = 1/f + 1/u

1/v = 1/f - 1/25

v = 25f / (f-25)

Now we can substitute this into the magnification formula and solve for f:

4 = -v/u

4 = -25f / u(f-25)

4 = -25f / (-25)(f-25)

4 = f / (f-25)

4f - 100 = f

3f = 100

f = 33.33 cm

Therefore, the focal length of the lens is approximately 33.33 cm.