Question

An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the size of the image.
(iii) Draw a ray diagram to show the formation of image in this case.​

Answer 1

Explanation:

(i)

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

u=-25cm

f=-15cm

$\frac{1}{v} + \frac{1}{ - 25} = \frac{1}{ - 15}$

$\frac{1}{v} - \frac{1}{25} = - \frac{1}{15}$

$\frac{1}{v} = - \frac{1}{15} + \frac{1}{25} \\ \frac{1}{v} = - \frac{5}{75} + \frac{3}{75} \: \: lcm = 75 \\ \frac{1}{v} = - \frac{2}{75} \\ v = - \frac{75}{2} \\ v = - 37.5cm$

(ii)

$m = \frac{hi}{ho} = \frac{ - v}{u} \\ \frac{hi}{4} = \frac{ - ( - 37.5)}{ - 25} \\ \frac{hi}{4} = \frac{ - 37.5}{25} \\ \frac{hi}{4} = - 1.5 \\ hi = - 1.5 \times 4 \\ hi = - 6 \: cm \\ therefore \: height \: or \: size \\ \: of \: the \: image \: is \: - 6 \: cm$

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