Question

An object is placed 27 cm in front of concave mirror of focal length 18 cm. At what distance from the mirror, should a screen be placed, so that a sharp focussed image can be obtained? Find the nature of the image. ​

Answer 1

$\huge{\underline{\underline{\underline{\sf{\pink{AnsWer \::-}}}}}}$

▬▬▬▬▬▬▬▬▬▬

To calculate the position of image

Here, object distance, u = - 27 cm

(to the left of mirror)

Focal length, f = - 18 cm

(it is a concave mirror)

Image distance, v = ?

(to be calculated)

We know mirror formula,

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

Now, putting these values in the mirror formula, we get,

$\frac{1}{v} + \frac{1}{ - 25} = \frac{1}{ - 18}$

or

$\frac{1}{v} = \frac{ - 1}{18} + \frac{1}{27}$

or

$\frac{1}{v} = \frac{ - 3 + 2}{54}$

or

$\frac{1}{v} = \frac{ - 1}{54}$

So, image distance, v = -54 cm

Thus, the image is formed at a distance of 54 cm in front of the concave mirror (minus sign shows the image formed on left side of the mirror). Therefore the screen should be placed at a distance of 54 cm in from of the concave mirror (on its left side).

Nature of image: Since the image is formed in front of the concave mirror, its nature will be real and inverted.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answer 2

Answer:

mark me as brainliest and support me to give more valuable answers

Explanation: