Question

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

Answer 1

Answer:

V=3.75,magnification= +0.25,Nature -virtual and errect

Answer 2

Image distance, u = -7.5 m

Size of the image, h' = -3 cm (inverted image)

Explanation:

Given that,

Height of the object, h = 6 cm

Object distance from a concave mirror, u = -15 cm

Focal length of the mirror, f = -5 cm

To find,

The image distance, magnification and nature of the image.

Solution,

Using mirror's formula to find the position of image as :

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

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

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

v = -7.5 cm

Let h' is the size of the image. Using formula of magnification to find it as :

$m=\dfrac{-v}{u}=\dfrac{h'}{h}$

$\dfrac{-(-7.5)}{(-15)}=\dfrac{h'}{6}$

h' = -3 cm

Negative sign shows that the formed image is inverted.  

Learn more,

Mirror's formula

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