Question

A screen having a real image of magnification m₁ formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m₂ is formed on the screen. The focal length of the lens is

• A. $\frac { x } { m _ { 2 } - m _ { 1 } }$
• B. $\frac { x } { m _ { 1 } - m _ { 2 } }$
• C. $\frac { x } { \sqrt { m _ { 1 } m _ { 2 } } }$
• D. None of these

Answer 1

Answer: (A)

$\frac { x } { m _ { 2 } - m _ { 1 } }$

Answer 2

In first case, $\frac { 1 } { p } + \frac { 1 } { q } = \frac { 1 } { f }$ and $\frac { q } { p }$ =m₁

⇒ 1+m₁ = $\frac { q } { f }$ . . . (1)

In the second case $\frac { 1 } { q + x } + \frac { 1 } { p ^ { \prime } } = \frac { 1 } { f }$ and $\frac { q + x } { p ^ { \prime } } = m _ { 2 }$

⇒ m₂ = $\frac { q + x } { f }$ . . . (2)

(1) and (2)

⇒ m₂ – m₁ = x/f ⇒ f = $\frac { x } { m _ { 2 } - m _ { 1 } }$ ∴