Question

The image of an object is formed on a screen at 1 1/2 of the distance of the object using a
concave mirror of focal length 9 cm. Find the object distance, image distance and
magnification
Answer ASAP
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Answer 1

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It is given that :

Image distance = 11/2u =? (given)

Object distance = u

Focal length = 9cm.

Magnification = m = ?

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Now according to mirror formula :

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

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

After inputting the given values in this equation we get :

$9 = \frac{ \frac{11u}{2} \times u}{ \frac{11u}{2} + u}$

$9 = \frac{11 {u}^{2} }{2} \div \frac{13u}{2}$

$9 = \frac{11 {u}^{2} }{13u} \\ \\ 9 = \frac{11u}{13}$

$13 \times 9 = 11u$

$u = \frac{117}{11} = 10cm$

Now we get Object distance is 10cm.

It is given that image distance is 11/2 times object distance. Therefore image distance :

$\frac{11}{2} \times10$

$= 55cm$

Now magnification :

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

After inputting the given values. :

$m = \frac{ - 55}{10}$

$m = - 5.5$

Thus magnification of image is 5.5.

(-) is due to the sign convention rule.

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BY SCIVIBHANSHU

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