Question

an object 5 cm in size is placed at 12 cm in front of concave mirror of radius of curvature 20 cm . At what distance from the mirror should or screen be placed in order to obtain a sharp image . find the nature and size of image .​​

Answer 1

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

It is given that :-

• Height of object = 5cm
• Object distance = u = 12cm
• Focal length of mirror = R /2 = 20/2 = 10cm.
• Image distance = v = ?
• Height of image = ?

Now, according to mirror formula :

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

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

After inputting the known values we get :

$\maltese \: 10 = \frac{ - 12v}{12 - v}$

$\maltese \: 10(12 - v) = - 12v$

$\maltese \: 120 - 10v = - 12v$

$\maltese \: 120 = - 12v + 10v$

$\maltese \: 120 = 2v$

$\maltese \: v = \frac{120}{2} = \cancel \frac{120}{2}$

$\maltese \: v = 60cm$

Thus a screen should be placed 60cm away from the mirror to obtain a sharp image.

Now we know that for a system of mirror the ratio of size or image to size of object = Ratio of image distance to object distance. Represented by :-

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

After inputting the known values we get :

$\frac{h_i}{5} = \frac{60}{12}$

$\frac{h_i}{5} = \cancel \frac{60}{12}$

$\frac{h_i}{5} = 5\mapsto \: h_i = 25cm$

Therefore we get that the size of image is 25cm.

Therefore we get that the size of image is 25cm.Now we can observe that size of object < Size of image, thus the image is magnified, real and inverted.

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BY SCIVIBHANSHU

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