Question

. Radius of curvature of a converging mirror is 40cm. An object is kept at 30cm from the mirror. Find the location and nature of the image.​

Answer 1

Focal length, f = radius of curvature / 2

==> 40 / 2

==> f = -20 cm ( because it is a converging mirror )

Object distance, u = -30 cm

We know,

==> 1/f = 1/v + 1/u

==> -1/20 = 1/v - 1/30

==> 1/v = 1/30 - 1/20

==> 1/v = 2 - 3/60

==> 1/v = -1/60

==> v = - 60 cm

Location of Image : 60 cm infront of the mirror

Nature of Image : Real , Inverted and Magnified.

Answer 2

Answer:

The image is real,inverted,enlarged and formed at a distance if 60m.

Explanation:

Given that,

The radius of curvature of a converging mirror is 40cm.

$= > R= - 40cm$

Converging mirror is nothing but an concave mirror.Also given that,an object is kept at an distance of 30cm from the mirror.As it is a concave mirror,the object distance is taken negative.

$= > u = - 30cm$

As radius is known,the focal length is

$f=\frac{R}{2} = \frac{ - 40}{2} = - 20cm$

Using the formula for mirrors to find the location of the image,we get

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

Substituting values of focal length and object distance in above relation,we get

$\frac{1}{ - 20} = \frac{1}{v} + \frac{1}{ - 30} \\ = > \frac{1}{v} = \frac{1}{30} - \frac{1}{20} = - \frac{1}{60} \\ = > v = - 60 \: cm$

Hence,the image is formed at a distance if 60cm from pole and it is real.Now the magnification of the image is

$m = \frac{ - v}{u} = \frac{ - ( -60)}{ - 30} = - 2$

As magnification is negative and greater than 1 the image is inverted and enlarged.

Therefore,The image is real,inverted,enlarged and formed at a distance if 60cm.

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