Question

in lens displacement method lateral magnification obtained are 2 and. 5. if separation between position of lenes is 30 CM then focal length of the lens is​

Answer 1

Answer:

-10cm.

Explanation:

Since, from the question we get that the magnification are given as 2 and 5. If we take the object distance is u and the image distance is v.

So, 2=v/u and 5=u/v.

Hence,2-5=v/u-u/v or -3=(v²-u²)/uv.

Which can be written as (v+u)(v-u)/uv. We also know that the separation is 30cm or v-u is 30.

Again, from the lense equation we get that 1/f=1/u+1/v from which we will get that (v+u)/uv will be 1/f.

So, on substituting the values we will get that:

-3=30/f.

Which on solving we will get the value of focal length as f=-10cm.