Question

for a concave mirror of focal length 20 centimetre if the object is at a distance of 30 cm from the pole than the nature of the image and magnification will be​

Answer 1

Explanation:

u = 30cm

f = 20cm

v=u.f/(u-f) = 30*20/(30-20)=600/10= 60cm

distance between object and image = v-u= 60-30= 30cm

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Answer 2

As per the data given in the above question.

we have to find the magnification and nature of the image .

Given,

Object distance, u = -30 cm

Focal length, f = -20 cm (concave mirror) Image distance, v =?

A concave mirror has a reflective surface that is curved inward and away from the light source. Concave mirrors reflect light inward to one focal point.

we know that ,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \: \: \: \: \: ....(1)$

Put the values in equation (1) ,

$\frac{1}{v} + \frac{1}{ - 30} = \frac{1}{ - 20}$

then shift the value to get the value of v,

$\frac{1}{v} = \frac{1}{ - 20} - \frac{1}{ - 30}$

Take L.C.M , and L.C.M is 60

$\frac{1}{v} = \frac{ - 3 + 2}{ 60}$

$\frac{1}{v} = \frac{ - 1}{ 60}$

$v = - 60 \: cm$

Now ,the magnification of the given terms

$m = \frac{ - v}{u}$

we have the values ,$v= -60 \: cm \: and \: u= -30 cm$

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

$m = \frac{ - 60}{ 30}$

$m = - 2$

Hence ,

The magnification is -2 and The image will be formed at infinity. The image formed in this case will be highly enlarged, real and inverted.

Project code # SPJ2