Question

11. An object 2.5 cm in length is placed at a distance of 10 cm in front of a convex
mirror of radius of curvature 15 cm. Find the position of the image, its nature and
size. Also draw its ray diagram.
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Answer 1

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

It is given that :-

• Height of object = 2.5cm
• Object distance = u = 10cm
• Focal length = R /2 = 15/2 = 7.5cm.
• Image distance = v = ?
• Height of Image = ?

Now according to mirror formula :

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

$\maltese \: f = \frac{uv}{u + v}$

After inputting the known values we get :

$\maltese \: 7.5 = \frac{ - 10v}{10 - v}$

$\maltese \: 7.5(10 - v) = - 10v$

$\maltese \: 75 - 7.5v = - 10$

$\maltese \: 75 = - 10v + 7.5v$

$\maltese \: 75 = 2.5v$

$\maltese \: v = \frac{75}{2.5} = \cancel \frac{750}{25} = 30cm$

Therefore we get, that the position of image will be 30cm away from the pole of mirror.

Now, we know that for a system of mirrors the ratio of size of image to size of object is equals to Ratio of Image distance to Object distance.

$\frac{h_i}{h_o} = \frac{v}{u}$

After inputting the known values we get :

$\frac{h_i}{2.5} = \frac{30}{10} = 3$

$h_i = 2.5 \times 3 = 7.5cm$

Therefore the size of image will be 7.5cm.

Now the image formed by a convex mirror is always same in nature that is virtual erect and diminished so it will also be same in this case.

The ray diagram for this system of mirrors is provided in the attachment. You may see it for understanding the concept more better.

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BY SCIVIBHANSHU

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