Question

an object is placed in front of concave mirror of focal length 15cm and its inverted sharp image of height 6cm is observed at a distance of 37.5cm from the mirror.find the object distance and its height​

Answer 1

Answer:mirror formula 1/f=1/v+1/u

1/u=1/f-1/v

f=15cm,v=37.5cm,u=?

Apply sign convention; f=-15cm,v=-37.5cm

1/u=1/-15-1/-37.5

1/u=-25+10/375

=-15/375

= -1/25

u=-25cm

magnification =hi/ho=-v/u

m= 6/ho=-37.5/25

ho=6/1.5

=4cm

Explanation:

Answer 2

The object distance is 25 cm and the height of the object is 4 cm.

Explanation:

Given that,

Focal length of the concave mirror, f = -15 cm

Height of the image, h' = 6 cm (inverted image)

Image distance, v = -37.5 cm (from the mirror)

The mirror's formula is given by :

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

$\dfrac{1}{(-15)}=\dfrac{1}{(-37.5)}+\dfrac{1}{u}$

u = -25 cm

On solving we get the value of object distance is (-25 cm). The formula of the magnification of mirror is given by :

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

$\dfrac{(-6)}{h}=\dfrac{-(-37.5)}{(-25)}$

h = 4 cm

So, the object distance is 25 cm and the height of the object is 4 cm. Hence, this is the required solution.

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