Question

An object 4cm tall is placed on the principal axis of a concave mirror of facal length 20 at a distance 30cm from it .find the position nature and size of the image.

Answer 1

Answer :-

✏ Position of image from mirror = -60 cm (infront of the mirror)

✏ Nature = Real, Inverted, enlarged and in front of the mirror

✏ Size of the image = -8 cm

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★ Concept :-

Here the concept Mirror Formula and Relationship of Magnification has been used.

• According to mirror formula,

$\bold{\dfrac{1}{f} \: = \: \bold{\dfrac{1}{v} \: + \: \dfrac{1}{u}}}$

where, f is the focal length, v is the position of image and u is the position of object.

• According to relationship of magnification of mirror,

$\bold{m} = \bold{\dfrac{h'}{h}} = \bold{\dfrac{(-v)}{u}}$

where, m is the magnification, h' is height of image, h is the height of object.

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★ Question :-

An object 4cm tall is placed on the principal axis of a concave mirror of focal length 20 at a distance 30cm from it . Find the position, nature and size of the image.

★ Solution :-

Given, let us apply the sign convention and see the information,

» Height of the object = h = 4 cm

» Focal length of the mirror = f = -20 cm

» Position of Image from the mirror = u = -30 cm

☯ Let the position of image from mirror be 'v'.

By applying the values, in mirror formula, we get,

$\bold{\dfrac{1}{f} \: = \: \bold{\dfrac{1}{v} \: + \: \dfrac{1}{u}}}$

$\bold{\dfrac{1}{(-20)}} \: = \: \bold{\dfrac{1}{v}} \: + \: \bold{\dfrac{1}{(-30)}}$

 $\bold{\dfrac{(-1)}{20}} \: = \: \bold{\dfrac{1}{v}} \: + \: \bold{\dfrac{(-1)}{30}}$

 $\bold{\dfrac{1}{v} = \bold{\dfrac{(-1)}{20}} + \bold{\dfrac{1}{30}}}$

On taking the LCM of denominators at RHS, we get,

$\bold{\dfrac{1}{v} = \bold{\dfrac{(-3 + 2)}{60}}}$

$\bold{\dfrac{1}{v}} = \bold{\dfrac{(-1)}{60}}$

On taking the reciprocal of the terms both the sides , we get

=> v = -60 cm

Hence, we get,

✒ Position of image = v = -60 cm

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Clearly, we see that the object is kept between centre of curvature and focal length of the mirror. So, the image formed will be behind centre of curvature of the mirror and before infinity. So,

✒ Nature = Real, Inverted, enlarged and in front of the mirror.

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Now let us find the size of the image. Then,

$\bold{m} = \bold{\dfrac{h'}{h}} = \bold{\dfrac{(-v)}{u}}$

 $\bold{\dfrac{h'}{h}} = \bold{\dfrac{(-v)}{u}}$

$\bold{h'} = \bold{\dfrac{(-v)h}{u}}$

By applying the values, in this equation, we get,

$\bold{h'} = \bold{\dfrac{-(-60)4}{(-30)}}$

✒  h' = -2 × 4

✒  h' = -8 cm

Here -ve sign shows that image is formed downwards that is in opposite direction to object.

So,

✒ Size of the image = h' = -8 cm

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★ More to know :-

• Sign Convention for Convex Mirror :-

• Position of image = +v
• Focal Length = +f
• Position of Object = -u

• While using mirror formula, the use of sign convention must be kept in mind properly for a better answer.

Answer 2

Answer:

• Position of image will be 60 cm infront of the mirror.
• Nature of image will be Real and inverted.
• Height (size) of image is -8 cm. ( negative sign shows that image formed is inverted )

Given:

• Height of object, $\sf{h_o}$ = 4 cm
• Focal length of concave mirror, $\sf{f}$ = -20 cm
• Position of object, $\sf{u}$ = -30 cm

To find:

• Position of image, $\sf{v}$ =?
• Height of image, $\sf{h_i}$ =?
• Nature of image

Formulae required

• Mirror formula

$\purple{\bigstar}\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

• Formula for magnification of mirror

$\purple{\bigstar}\boxed{\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}}}$

[ Where v is position of image, u is position of object, f is focal length of mirror, m is magnification, $\sf{h_i}$ is position of image and $\sf{h_o}$ is position of object ]

Solution

Using Mirror formula to calculate position of image

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

$\implies\sf{\dfrac{1}{-20}=\dfrac{1}{v}+\dfrac{1}{(-30)}}$

$\implies\sf{\dfrac{1}{-20}=\dfrac{1}{v}-\dfrac{1}{30}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{1}{-20}+\dfrac{1}{30}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{-3+2}{60}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{-1}{60}}$

$\underline{\underline{\implies\purple{\sf{v=-60\;cm}}}}$

Calculating height of image using formula for magnification of mirror

$\implies\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}}$

$\implies\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o}}$

$\implies\sf{\dfrac{-(-60)}{-30}=\dfrac{h_i}{4}}$

$\implies\sf{\dfrac{60}{-30}\times 4=h_i}$

$\underline{\underline{\implies\sf{\purple{h_i=-8\;\;cm}}}}$

Therefore,

• Position of image will be 60 cm infront of the mirror.
• Nature of image will be Real and inverted.
• Height (size) of image is -8 cm. (negative h$\sf{{}_i}$ because image formed in inverted)