The distance between an object and a divergent lens is m times the focal length of the lens. The linear magnification produced by the lens is:

Question

The distance between an object and a divergent lens is m times the focal length of the lens. The linear magnification produced by the lens is: (A) m (B) 1 / m (C) m+1 (D) (1/m+1)

Tardigrade Admin · Accepted Answer

Here u = mf ∴ for divergent lens (concave) ∴ v =( m f / m +1) Now magnification =( v / u ) ⇒ m =( m f / m +1) × (1/ mf ) m =(1/ m +1)

Q. The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is:

m

$1/ m$

m+1

$\frac{1}{m + 1}$

Here $u = m f$
$∴$ for divergent lens (concave)
$∴ v = \frac{m f}{m + 1}$
Now magnification $= \frac{v}{u}$
$\Rightarrow m = \frac{m f}{m + 1} \times \frac{1}{m f}$
$m = \frac{1}{m + 1}$

Q. The distance between an object and a divergent lens is m times the focal length of the lens. The linear magnification produced by the lens is:

m

1/m

m+1

m+11​

Here u=mf ∴ for divergent lens (concave) ∴v=m+1mf​ Now magnification =uv​ ⇒m=m+1mf​×mf1​ m=m+11​

Q. The distance between an object and a divergent lens is $m$ times the focal length of the lens. The linear magnification produced by the lens is:

$1/ m$

$\frac{1}{m + 1}$

Here $u = m f$
$∴$ for divergent lens (concave)
$∴ v = \frac{m f}{m + 1}$
Now magnification $= \frac{v}{u}$
$\Rightarrow m = \frac{m f}{m + 1} \times \frac{1}{m f}$
$m = \frac{1}{m + 1}$