Question

2.If an object is placed at a distance of 5 cm before
a concave mirror of focal length 8 cm, hen write the
characteristics of image?​

Answer 1

Answer:

Explanation:

here , given

u = -5 cm

f = - 8 cm

by, mirror formula

1/f = 1/v + 1/u

1/-8 = 1/v + 1/-5

⇒ 1/v =  1/5 - 1/8

 ∴ , 1/v = 8 - 5/40

    1/v = 3/40

v = 40/3 cm

v =  13.33333333333333333333333333333333333333333333...... cm

⇒ v ≈ 13.34 cm

but,

m= - v / u

m = -13.34 / -5

m = 2.668

therefore the properties of image

it is a virtual and erect image , which is 2.668 times larger than that of object size and is formed 13.34 cm behind the concave mirror.

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Answer 2

Given:

• Position of Object, u = - 5 cm
• Focal length of concave mirror, f = -8 cm

To find:

• Characteristics of image

1. Position of image, v =?
2. Magnification of image, m =?
3. Nature of Image

Formula required:

• Mirror formula

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

• Formula for magnification produced by spherical Mirrors

$\purple{\star}\boxed{\sf{m=\dfrac{-v}{u}}}$

[ Where f is focal length of mirror, v is position of image, u is position of object, m is magnification ]

Solution:

Using mirror formula  

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

$\implies\sf{\dfrac{1}{-8}=\dfrac{1}{-5}+\dfrac{1}{v}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{1}{-8}+\dfrac{1}{5}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{-5+8}{40}}$

$\implies\sf{\dfrac{1}{v}=\dfrac{3}{40}}$

$\implies\underline{\underline{\purple{\sf{v=\dfrac{40}{3}=13.34\;\;cm}}}}$

Using formula for magnification of Mirror

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

$\implies\sf{m=\dfrac{-40/3}{-5}}$

$\implies\sf{m=\dfrac{40}{15}}$

$\implies\underline{\underline{\purple{\sf{m=2.67}}}}$

Therefore,

• Position of image is at 13.34 cm behind the mirror.
• Magnification of image is 2.67 therefore, image formed would be enlarged.
• Nature of image would be virtual and erect.