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

Question

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

Tardigrade Admin · Accepted Answer

The given lens is a convex lens. Let the magnification be m, then for real image using lens formula

\frac{1}{m x} + \frac{1}{x} = \frac{1}{f}

.......(i)
And for virtual image

\frac{1}{- m y} + \frac{1}{y} = \frac{1}{f}

........(ii)
From equation (i) and equation (ii),
we get

f = \frac{x + y}{2}

Q. When an object is at distance $x$ and $y$ from a lens, a real image and a virtual image are formed respectively having the same magnification. The focal length of the lens is

$\frac{x + y}{2}$

$x - y$

$x y$

$x + y$

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

2x+y​

x−y

xy​

x+y

The given lens is a convex lens. Let the magnification be m, then for real image using lens formula mx1​+x1​=f1​ .......(i) And for virtual image −my1​+y1​=f1​ ........(ii) From equation (i) and equation (ii), we get f=2x+y​

Q. When an object is at distance $x$ and $y$ from a lens, a real image and a virtual image are formed respectively having the same magnification. The focal length of the lens is

$\frac{x + y}{2}$

$x - y$

$x y$

$x + y$