Question

a short linear object of length 3cm lies along the axis of a concave mirror of focal length 15cm. The midpoint of the object is at distance of 26.5 cm from the pole of the mirror. The size of the image is equal to ?​

Answer 1

Answer:

5.1

Explanation:

Answer 2

Size of image  is equal to 3.9 cm

Explanation:

Length of object=3 cm

Focal length of concave mirror=f=-15 cm

Object distance =-26.5 cm

Mirror formula

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

Substitute the values in the formula

Then, we get

$-\frac{1}{15}=\frac{1}{v}-\frac{1}{26.5}$

$\frac{1}{v}=-\frac{1}{15}+\frac{1}{26.5}$

$\frac{1}{v}=\frac{-26.5+15}{397.5}$

$\frac{1}{v}=-\frac{11.5}{397.5}$

$v=-\frac{397.5}{11.5}=-34.57$ cm

Magnification=$- \frac{v}{u}$

Magnification=$- \frac{34.57}{26.5}=-1.3$

Again magnification formula for concave mirror

$m=\frac{h'}{h}$

Where h'=Height of image

h=Height of object

Using the formula

Then, we get $\frac{h'}{3}=-1.3$

$h'=3\times (-1.3)=-3.9 cm$

Hence, size of image is equal to 3.9 cm

Where negative sign shows that image is inverted.

#Learns more:

https://brainly.in/question/9168419