Question

An object is placed at a distance of 60cm from a convex mirror where the
magnification produced is 1/2. Where the object should be placed to get a magnification of 1/3 ?​

Answer 1

Given :-

▪ An Object is placed at a distance of u = - 60 cm

from a convex mirror. The magnification produced is 1/2

To Find :-

▪ Distance at which the object should be put in order to get a magnification of 1/3.

Solution :-

The magnification is 1/3 which is given by,

⇒ m = -v / u

⇒ 1/2 = -v/-60

⇒ -60/2 = -v

⇒ v = +30 cm

Now that we have found the image and object distance, focal length of the mirror can be calculated as

⇒ 1/f = 1/u + 1/v

⇒ 1/f = -1/60 + 1/30

⇒ 1/f = (-1 + 2)/60

⇒ 1/f = 1/60

⇒ f = +60 cm

We had to find the object distance when the magnification is 1/3, so

⇒ m = -v / u

⇒ 1/3 = -v/u

⇒ v = - u / 3

⇒ 1/v = -3/u

Using the mirror formula, we have

⇒ 1/f = 1/v + 1/u

⇒ 1/60 = -3/u + 1/u

⇒ 1/60 = (-3 + 1)/u

⇒ 1/60 = -2/u

⇒ u = -120 cm

Hence, The object should be kept at a distance of 120 cm infront of the mirror.

Answer 2

➣Given:

• Object Distance, u = -60cm
• Magnification, m = 1/2 = 0.5

➣Find:

• Where should the object be placed

➣Solution:

we, know

Magnification is given by

$\sf \to m = - \frac{v}{u}$

Putting the values, we get

$\sf \to 0.5 = - \frac{v}{ - 60}$

$\sf \to - v = (0.5)( - 60)$

$\sf \to - v = - 30$

$\underline{ \boxed{ \sf \therefore v = 30cm}}$

___________________

Now, using mirror formula,

$\sf \to \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

Putting the values we get,

$\sf \to \frac{1}{ - 60} + \frac{1}{30} = \frac{1}{f}$

$\sf \to \frac{ - 1 + 2}{60} = \frac{1}{f}$

$\sf \to \frac{ 1}{60} = \frac{1}{f}$

$\underline{\boxed{\sf \therefore f = 60cm}}$

______________________

From, questioning for magnification, m = 1/3

Using the formula,

$\sf \to m = - \frac{v}{u}$

Putting the values we get,

$\sf \to \frac{1}{3} = - \frac{v}{u}$

$\underline{ \boxed{ \sf \therefore v = - \frac{u}{3}}}$

______________________

Now, by using mirror formula

$\sf \to \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

Putting the values we get,

$\sf \to \frac{1}{u} + \frac{1}{( - \frac{u}{3} )} = \frac{1}{60}$

$\sf \to \frac{1}{u} - \frac{3}{u} = \frac{1}{60}$

$\sf \to \frac{1 - 3}{u} = \frac{1}{60}$

$\sf \to \frac{ - 2}{u} = \frac{1}{60}$

$\sf \to u = ( - 2)(60)$

$\underline{\boxed{\sf \therefore u = - 120cm}}$

Hence, the required object distance = 120cm