Question

A convex mirror of focal length 10 cm. A object is placed 12cm away from pole find its img.... distance and magnification​

Answer 1

Answer: This may help you

Explanation:

It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.

Given: Object distance, u=−10cm Focal length, f=15cm

To find: Image distance, v

From mirror formula :

v

1

​

+

u

1

​

=

f

1

​

v

1

​

−

10

1

​

=

15

1

​

v

1

​

=

15

1

​

+

10

1

​

=

150

25

​

v=6cm

Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification

m=

u

−v

​

=

−10

−6

​

=0.6

Hence it will be erect and diminished.

Answer 2

The distance of the image is 5.45 cm

The magnification of the image is 5/11

The image will be virtual, erect and diminished

Explanation:

Distance of the object u = 12 cm

Focal length of the mirror f = -10 cm  (Since it is convex lens)

From the mirror formula if the distance of the image is v then

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

$\implies \frac{1}{v}+\frac{1}{12}=\frac{1}{-10}$

$\implies \frac{1}{v}=-(\frac{1}{10}+\frac{1}{12})$

$\implies \frac{1}{v}=-\frac{6+5}{60}$

$\implies\frac{1}{v}=-\frac{11}{60}$

$\implies v=-\frac{60}{11}$ cm  

$\implies v=-5.45$ cm   (negative side implies that the image is on the other side of the object i.e. the image is virtual)

The magnification is given by

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

$\implies m=-\frac{-60/11}{12}$

$\implies m=\frac{5}{11}$

The image will be erect and diminished

Hope this answer is helpful.

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