When an object is at a distance u1 and u2 from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

Question

When an object is at a distance u1 and u2 from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is (A) u1-u2 (B) (u1-u2/2) (C) (u1+u2/2) (D) u1+u2

Tardigrade Admin · Accepted Answer

m=(f/f+u) For real image For virtual image -m=(f/f-u1), +m=(f/f-u1), -(f/f-u2)=(f/f-u1) ⇒ f=(u1+u2/2)

Q. When an object is at a distance $u_{1}$ and $u_{2}$ from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

$u_{1} - u_{2}$

$\frac{u _{1} - u _{2}}{2}$

$\frac{u _{1} + u _{2}}{2}$

$u_{1} + u_{2}$

$m = \frac{f}{f + u}$
For real image
For virtual image
$- m = \frac{f}{f - u _{1}}, + m = \frac{f}{f - u _{1}},$
$- \frac{f}{f - u _{2}} = \frac{f}{f - u _{1}}$
$\Rightarrow f = \frac{u _{1} + u _{2}}{2}$

Q. When an object is at a distance u1​ and u2​ from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

u1​−u2​

2u1​−u2​​

2u1​+u2​​

u1​+u2​