Question

An object 4 cm in height is placed at 15 cm in front of a concave mirror o focal length 10 cm at what distance from the mirror should a screen be placed to obtain a sharp image of the object calculate the Height of image.​

Answer 1

Answer:

• For obtaining sharp image of the object, screen should be placed at a distance of 30 cm in front of mirror.
• Height of image is 8 cm

Explanation:

Given :

• $\sf h_{o} = + 4 \: cm$
• $\sf u = - 15 \: cm$
• $\sf f = -10 \: cm$

To find :

• Image distance, v.
• Height of image, $\sf h_{i}$.

Solution :

By Mirror formula :

• $\sf \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

[Where, f is focal length, v is image distance and u is object distance]

$\longrightarrow \sf \dfrac{1}{-10} = \dfrac{1}{v} + \dfrac{1}{-15}$

$\longrightarrow \sf \dfrac{1}{-10} + \dfrac{1}{15} = \dfrac{1}{v}$

$\longrightarrow \sf \dfrac{(-15 + 10)}{150} = \dfrac{1}{v}$

$\longrightarrow \sf \dfrac{-5}{150} = \dfrac{1}{v}$

$\longrightarrow \sf \dfrac{-1}{30} = \dfrac{1}{v}$

$\longrightarrow \pmb{\sf -30 = v}$

Or, v = -30

For obtaining sharp image of the object, screen should be placed at a distance of 30 cm in front of mirror.

Now,

By magnification:

• $\sf m = \dfrac{-v}{u} = \dfrac{h_{i}}{h_{o}}$

So,

$\longrightarrow \sf m = - \bigg(\dfrac{-30}{-15}\bigg)$

$\longrightarrow \pmb{\sf m = -2}$

And,

$\longrightarrow \sf -2 = \dfrac{h_{i}}{4}$

$\longrightarrow \sf -2 \times 4 = h_{i}$

$\longrightarrow \pmb{\sf h_{i} = -8}$

Therefore, Height of image is 8 cm.

Answer 2

Given :-

An object 4 cm in height is placed at 15 cm in front of a concave mirror o focal length 10 cm

To Find :-

Distance from the mirror should a screen be placed to obtain a sharp image of the object calculate the Height of image.​

Solution :-

We know that

1/u + 1/v = 1/f

1/-15 + 1/v = 1/-10

-1/15 + 1/v = -1/10

1/v = -1/10 + 1/15

1/v = (-3 + 2)/30

1/v = -1/30

v = -30 cm

Now

m = -v/u = hi/ho

-(-30)/-15 = hi/4

30/-15 = hi/4

-2/1 = hi/4

-8 = hi

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