Question

An object is height of 6cm is placed perpendicular on the principal axis before concave mirror at the distance of 15cm the focal length of the concave mirror is 5cm find the image distance magnification and nature of image

Answer 1

Answer:

Image will form at 7.5 cm from the mirror in front of it

Magnification of image is 1.5 times

Nature of image is real, Inverted and magnified

Explanation:

As we know that the formula for the mirror is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know that

$d_o = 15 cm$

$f = 5 cm$

now we have

$\frac{1}{d_i} + \frac{1}{15} = \frac{1}{5}$

$\frac{1}{d_i} = \frac{1}{5} - \frac{1}{15}$

$d_i = 7.5 cm$

now we have

$magnification =-\frac{d_i}{d_o}$

$M = -\frac{7.5}{5}$

$M = -1.5$

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Topic : Curved Mirror

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Answer 2

The image is located at a distance of 7.5 cm behind the mirror and the image is real and inverted.

Explanation:

Given that,

Height of the object, u = 6 cm

Object distance, u = -15 cm

Focal length of the concave mirror, f = -5 cm (it is negative for concave mirror)

Let v is the position of the image with respect to the object. We can find the image distance using mirror's formula as :

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

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-5}-\dfrac{1}{-15}$

v = -7.5 cm

The magnification of the mirror is given by :

$m=\dfrac{-(-7.5)}{-15}$

m = -0.5

So, the image is located at a distance of 7.5 cm behind the mirror and the image is real and inverted. Hence, this is the required solution.

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Mirror's formula