Question

an object 6 CM in size is placed at 30 cm in front of a concave mirror of focal length 15 CM at what distance from the mirror should a object be placed in order to obtain a sharp image? find the nature and size of the image.​

Answer 1

Answer:

v= -50cm

Explanation:

By sign convention:

Object Distance,u: -30cm

Focal length,f:-15cm

Image distance,v:?

Applying Mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \\ \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ \\ \implies \: \frac{1}{v} = \frac{1}{ - 25} - \frac{1}{ - 50} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 1}{25} + \frac{1}{50} = \frac{ - 2 + 1}{50} \\ \\ \\ \implies \: \frac{1}{v} = \frac{ - 1}{50} \\ \\ \\ \implies \: \boxed{v = - 50cm}$

Characteristics of image formed:

1)Real and inverted

2)Same size as the object

3)Formed in front of the mirror

Now,

•Magnification factor,m:

$m = \frac{ - v}{u} = - \frac{ - 50}{ - 50} = - 1$

Height of object:6cm

Height of image:?

Also,

m= h`/h

=> -1=h`/6

=>h`= -6

Height of the image is -6