Question

If $m_1$ and $m_2$ be the linear magnification of the object in the conjugate positions of a convex lens, and if d be the distance between the conjugate positions, then the focal length of the lens is given by:

• A. $f=\frac{d}{m_1-m_2}$
• B. $f=\frac{d}{m_1+m_2}$
• C. $f=\frac{m_1-m_2}{d}$
• D. either (a) and (b)

Answer 1

Answer: (A)

For the first position $m_1=\frac{\upsilon }{u}$
For the second position $m_2=\frac{u}{\upsilon }$ From the lens formula $f=\frac{u\upsilon }{u+\upsilon }$ or $f=\frac{u\upsilon }{u+\upsilon }\times \frac{\upsilon -u}{\upsilon -u}$ $=\frac{(\upsilon -u)}{({\upsilon }^2-u^2)/u\upsilon }$ i.e., $f=\frac{u-\upsilon }{\left(\frac{\upsilon }{u}-\frac{u}{\upsilon }\right)}$ But $\upsilon -u=d$ $\therefore$ $f=\frac{d}{m_1-m_2}$