Question

An object 1cm high,is placed on the principal axis of a concave mirror at a distance of 5cm from the mirror if the focal length is 10cm.Calculate the image distance and its magnification.

Answer 1

Given :

• Focal length (f) = - 12cm

• Object distance (u) = - 18cm

To Find :

• Image distance

• Magnification

Solution :

★ Applying Mirror Formula

$:\implies\tt \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f} \\ \\ \\ :\implies\tt \dfrac{-1}{18}+\dfrac{1}{v}=\dfrac{-1}{12} \\ \\ \\ :\implies\tt \dfrac{1}{v}=\dfrac{1}{18}-\dfrac{1}{12} \\ \\ \\ :\implies\tt \dfrac{1}{v}=\dfrac{2-3}{36} \\ \\ \\ :\implies\tt \dfrac{1}{v}=\dfrac{-1}{36} \\ \\ \\ :\implies\sf v=-36cm$

Hence, image distance is - 36 cm

we know that,

$:\implies\tt m=\dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ :\implies\tt m=\dfrac{-v}{u} \\ \\ \\ :\implies\tt m=\dfrac{-(-36)}{(-18)} = -2$

Hence, magnification is - 2

Answer 2

Given:-      

• Focal length (f) = - 12cm
• Object distance (u) = - 18cm

To find:-    

• Image distance
• Magnification

Solution:-    

~Image distance

• To find out the image distance we will use the mirror formula ( 1/u + 1/v = 1/f ).

➳  1/u + 1/v = 1/f

➳ -1/18 + 1/v = -1/12

➳  1/v = 1/18 - 1/12

➳  1/v = 2 - 3/36

➳  1/v = -1/36

. ° . Thus, The image distance is -36 cm.

               ___________________

~Magnification

• To find out magnification we will use the magnification formula ( Magnification = -v/u )

➳ Magnification = -v/u = hi/ho

➳ Magnification = -v/u

➳ Magnification = - ( -36 ) / ( - 18 )

➳ Magnification = -2

. ° . Thus, The magnification is -2 cm.