UPSC CDS Science Physics Questions Part 1

(c) Let s be the distance of object and image, u be the object distance and f be the focal length of the lens.
Then, the lens formula, ‌‌

1

u

+‌

1

s−u

‌=‌

1

f

⇒f(s−u)+fu−u(s−u)‌=0
⇒fs−fu+fu−su+u2‌=0
⇒‌fs−su+u2=0
⇒‌u2=(u−f)s
‌ or ‌‌s=‌

u2

u−f

On differentiating,
‌

ds

du

‌‌=‌

(u−f)(2u)−u2

(u−f)2

‌‌=‌

(u2−2uf)

(u−f)2

=‌

u(u−2f)

(u−f)2

‌

ds

du

‌‌=0,‌ when ‌u(u−2f)=0
⇒‌‌u‌‌=0
For real image, u=2f
When u<2f,‌

ds

du

<0.
When u>2f,‌

ds

du

>0
u=2f is therefore a minimum. The
object and image distances for
minimum separation are both 2f, giving
separation 4f.