Question

An object 4cm in size is placed at a distance of 25 cm from a concave mirror of focal length 15 cm. find the position, nature, and the height of the image.

Answer 1

Answer:

nullk hom 258 B ecause it is the whole length

Answer 2

GIVEN:-

•Height of the object ,$\sf{h_1}$=+4m

•Object distance ,u=-25.0cm

•Focal length of concave mirror, f=-15.0cm

TO FIND OUT:-

•Position of the image

•Nature of the image

•Height of the image

SOLUTION:-

•Position of the image

Using mirror formula

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

Or $\sf\large\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\sf\implies\large \frac{1}{v}=\frac{1}{-15}-\frac{1}{-25}$

$\sf\large\implies \frac{1}{v}=\frac{-5+3}{75}$

$\sf\large\implies v=\frac{75}{-2}=-37.5cm$

i.e, Image distance ,v=-37.5cm

•Nature of the image

Since the image is formed in front of the mirror, because v is negative

•Height of the image

•Using expression for linear magnification

$\sf\large m=\frac{h_2}{h_1}=\frac{-v}{u}$

And putting $\sf h_1$=+4cm ,u=-25.0cm and v=-37.5cm,We get

$\sf\large\frac{h_2}{4}=-\bigg(\frac{-37.5}{-25.0}\bigg)=\frac{-3}{2}$

$\sf \large h_2=-6cm$

Therefore,height of the image is 6 cm .The negative sign shows that the image is below the principle axis.Therefore,it is inverted and real