Question

an object is placed at 10 centimetre form a concave mirror produces three times magnified real image calculate focal length of the mirror​

Answer 1

Answer :-

Focal length of the mirror is 15 cm .

Explanation :-

We have :-

→ Position of the object = 10 cm

→ Image is real and 3 times magnified.

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Since, the mirror is concave :-

• u = -10 cm

By the formula of magnification, we have :-

m = -(v/u)

⇒ -(v/-10) = 3

⇒ v = 10(3)

⇒ v = 30 cm

Now, according to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/f = 1/30 + 1/(-10)

⇒ 1/f = 1/30 + (-1)/10

⇒ 1/f = (1 - 3)/30

⇒ 1/f = -2/30

⇒ -2f = 30

⇒ f = 30/-2

⇒ f = -15 cm

[We have got value of 'f' as -ve, as focal length is always -ve for a convave mirror.]

Answer 2

Answer:

Given :-

• An object is placed at 10 cm from a concave mirror produces three times magnified real image.

To Find :-

• What is the focal length of the mirror.

Formula Used :-

$\clubsuit$ Magnification Formula :

$\longmapsto \sf\boxed{\bold{\pink{Magnification\: (m) =\: \dfrac{- v}{u}}}}$

where,

• v = Image Distance
• u = Object Distance

$\clubsuit$ Mirror Formula :

$\longmapsto \sf\boxed{\bold{\pink{\dfrac{1}{v} + \dfrac{1}{u} =\: \dfrac{1}{f}}}}$

where,

• v = Image Distance
• u = Object Distance
• f = Focal Length

Solution :-

First, we have to find the image distance :

Given :

• Magnification = 3
• Object Distance = - 10 cm

According to the question by using the formula we get,

$\implies \sf 3 =\: \dfrac{\cancel{-} v}{\cancel{-} 10}$

$\implies \sf 3 =\: \dfrac{v}{10}$

By doing cross multiplication we get,

$\implies \sf v =\: 10(3)$

$\implies \sf v =\: 10 \times 3$

$\implies \sf\bold{\green{v =\: 30\: cm}}$

Hence, the image distance is 30 cm .

Now, we have to find the focal length :

Given :

• Object Distance = - 10 cm
• Image Distance = 30 cm

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{1}{30} + \bigg(- \dfrac{1}{10}\bigg) =\: \dfrac{1}{f}$

$\longrightarrow \sf \dfrac{1}{30} - \dfrac{1}{10} =\: \dfrac{1}{f}$

$\longrightarrow \sf \dfrac{1 - 3}{30} =\: \dfrac{1}{f}$

$\longrightarrow \sf \dfrac{- 2}{30} =\: \dfrac{1}{f}$

By doing cross multiplication we get,

$\longrightarrow \sf -2f =\: 30(1)$

$\longrightarrow \sf - 2f =\: 30$

$\longrightarrow \sf f =\: \dfrac{\cancel{30}}{- \cancel{2}}$

$\longrightarrow \sf \bold{\red{f =\: - 15\: cm}}$

$\therefore$ The focal length of the mirror is - 15 cm.