Physics, asked by nihal3378, 8 months ago

a 2 cm object is placed at a distace of 20 cm from a concave mirror a real image is formed 40 cm from the mirror. calculate the focal length of the mirror and size of the image​

Answers

Answered by VishalSharma01
51

Answer:

Explanation:

Solution,

Here, we have

Height of object = 2cm

Object distance, u = - 20 cm

Image distance, v = - 40 cm

To Find,

Focal length, f = ?

Height of image = ?

According to the mirror formula,

We know that,

1/f = 1/v + 1/u

So, putting all the values, we get

1/f = 1/v + 1/u

⇒ 1/f = 1/- 40 + 1/- 20

⇒ 1/f = 1/- 40 - 1/20

⇒ 1/f = - 1 - 2/40

⇒ 1/f = - 3/40

⇒ f = - 40/3

f = - 13.33 cm

Hence, the focal length is - 13.33 cm.

Now, the Height of the image,

According to the magnification formula,

We know that,

Magnification, m = h(i)/h(o) = - v/u

So, putting all the values, we get

⇒ h(i)/h(o) = - v/u

⇒ h(i)/2 = -(- 40/- 20)

h(i) = - 4 cm

Hence, the height of image is - 4 cm.

Answered by rohit301486
1

Given:

a 2 cm object is placed at a distace of 20 cm from a concave mirror a real image is formed 40 cm from the mirror.

To find:

calculate the focal length of the mirror and size of the image

STEP BY STEP EXPLANATION:

Object distance (u) = -20cm

Image distance (v) = -40cm

Height of object ( {h}^{o} ) = 2cm

Mirror formula

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

 \frac{1}{ - 40}  +  \frac{1}{ - 20}  =  \frac{1}{f}  \\

 \frac{ - 1 - 2}{40} =  \frac{1}{f}  \\

f = \ -  \frac{40}{3}  =  - 13.33 \: cm \\

Magnification, m =  \frac{ - v}{u}  =  \frac{ {h}^{i} }{ {h}^{o} }  \\ </p><p></p><p>\ - \frac{ - 40}{ - 20}  =  \frac{ {h}^{i}  }{2}  \\

the size of the image is  {h}^{i}  =  - 4cm

Hence verified !

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