Question

how far should an object be held from a concave mirror of focal length 40 cm so as to get an image magnified three times

Answer 1

Answer:

Given:-

Focal length, (f) = -40 cm

Magnification, (m) = 3

To Calculate:-

Distance of the object, (u) = ???

Since the nature of the image so formed is not specified, so two cases arise here.

Case 1 : When the image formed is real.

Case 2 : When the image formed is virtual.

Formula to be used:-

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

Since only f and m are given, therefore, we cannot use this formula directly and instead first use the formula,

⇒ $m=\frac{v}{u}$

⇒ $3=\frac{v}{u}$

⇒ $v=3u$

Case 1: When the image formed is real,

⇒ $v=+3u$

Substituting the given values, we get

⇒ $\frac{1}{3u}+\frac{1}{u}=\frac{1}{-40}$

⇒ $\frac{1+3}{3u}=\frac{1}{-40}$

⇒ $\frac{3u}{u}-40$

⇒ $u=-53.33 \: cm$

Case 2 : When the image formed is virtual.

⇒ $v=-3u \: cm$

Substituting the given values, we get

⇒ $-\frac{1}{3u}+\frac{1}{u}=-\frac{1}{40}$

⇒ $\frac{-1+3}{3u}=-\frac{1}{40}$

⇒ $\frac{2}{3u}=-\frac{1}{40}$

⇒ $-80=3u$

⇒ $\frac{-80}{3}=u$

⇒ $u=-26.67 \: cm$

Hence, if the object is held at a distance of 53.33 cm, a real image thrice in size is obtained.

Also, a virtual image of thrice the size is obtained if the distance of the object is 26.67 cm.

Answer 2

Answer:

-26. 67cm

Explanation:

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