Question

the focal length of a concave mirror is 10cm where should an object be placed so as to get its real image magnified 4 times
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Answer 1

★Given:-

• Focal length of a concave mirror = 10cm
• Magnification of it's real image = 4 times

★To find:-

• Position of object.

★Solution:-

We know:

✦Magnification = -v/u

Where,

• v = image distance
• u = object distance

Putting values,

⇒m = -v/u

⇒4 = -v/u

⇒v = -4u

Using the formula,

✦1/f = 1/v + 1/u

Where,

• f = focal length = -10cm
• v = image distance = -4u
• object distance = -u

Putting values in the formula,

⇒1/(-10) = 1/(-4u) + 1/(-u)

⇒1/(-10) = 1/(-4u) - 1/u

⇒1/(-10) = u - (-4u) / -4u²

⇒1/(-10) = u + 4u / -4u²

⇒1/(-10) = 5/-4u

⇒1/10 = 5/4u

⇒4u = 50

⇒u = 50/4

⇒u = 12.5cm

Hence, the object should be placed at 12.5cm.

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Answer 2

$\huge{\textsf{\underline{\underline{Question:-}}}}$

The focal length of a concave mirror is 10cm where should an object be placed so as to get its real image magnified 4 times.

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$\huge{\textsf{\underline{Answer:-}}}$

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${\sf{\underline{Given:}}}$

▪︎Focal length of concave mirror = 10cm

▪︎Magnification of real image = 4 times

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${\sf{\underline{To\ Find:}}}$

▪︎Position of object

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${\sf{\underline{Solution:}}}$

According to Formula;

Magnification = $\frac{-v}{u}$

Where value of

v = image distance

u = object distance

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Now,

${\bf{\implies m = -v/u}}$

${\bf{\implies 4 = -v/u}}$

${\bf{\implies v = -4u}}$----Equation①

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As we know;

According to Formula

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

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By Putting Equation① in given formula

$\sf \frac{1}{-10} = \frac{1}{-4u} +\frac{1}{u}$

$\implies \frac{1}{-10} = \frac{u-(-4u)}{-4u²}$

$\implies \frac{1}{-10} = \frac{5u}{4u²}$

$\implies \frac{1}{\cancel-10} = \frac{5 \cancel{u}}{\cancel-4u× \cancel{u}}$

$\implies \frac{1}{10} = \frac{5}{4u}$

$\implies \frac{1}{10} = \frac{5u}{4u²}$

$\implies 4u = 50$

$\implies u = 12.5$

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$\therefore Object\ should\ be\ placed\ 12.5\ cm\ away$