Question

where should an object be placed in front of a convex lens of focal length 40 cm,so as to obtain a real and inverted image magnified 4 times.​

Answer 1

Given:

The focal length of the convex lens = 40 cm

Magnification = -4 ( Image is inverted )

To Find:

We have to find the object distance from the optic center for an image placed in front of the convex lens of focal length 40 cm.

Solution:

The magnification factor means how large an object or subject can be reproduced on the image. That is, it is the ratio of the height of the image to the height of the object. Also, it can be expressed as the ratio of the image distance to the object distance.

That is, m = $\frac{u}{v}$

where

m ⇒ magnification factor

u ⇒ object distance from the optic center

v ⇒ image distance from the optic center

For an inverted image, m is negative.

                             m = $\frac{u}{v}$

                     ⇔   -4 = $\frac{u}{v}$

                     ⇒    -u = 4v     ...............Eq. 1

The lens formula is given by,

                    $\[\frac {1}{f} = \frac{1}{v} - \frac{1}{u}\]$

Substitute Eq. 1 and value of 'f' in the above formula, then we get

                    $\[\begin{array}{l}\frac{1}{{40}} = \frac{1}{v} - \frac{1}{{4v}}\\\\\frac{1}{{40}} = \frac{3}{{4v}}\\\\v = 30\,cm\end{array}\]$

Hence, we get the image distance as 30cm

⇒ The object distance, -u = 4 × 30 = 120 cm

    ∴ u = -120 cm

∴ The object has to be placed at 120 cm away in front of the mirror.

                       

Answer 2

Answer:

Here,focal length of the convex lens, f = +4

As magnification is 4 and the image is real and inverted, hence m = -4

m =v/u

v=mu=(-4)u = -4u

from the lens formula,1/v - 1/u = 1/f

1/(-4u) - 1/(u) = 1/40

-1 - 4/4u = 1/40

-5/4u = 1/40

u=-5×40/4

u=-50 cm ANS