Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance
[tex]u -v= d[/tex]
[tex]u=71+26=97\ cm[/tex]
We need to calculate the focal length
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
put the value into the formula
[tex]\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}[/tex]
[tex]\dfrac{1}{f}=-\dfrac{71}{2522}[/tex]
[tex]f=-35.52\ cm[/tex]
The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value in to the formula
[tex]\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{124}{2755}[/tex]
[tex]v=-22.21\ cm[/tex]
We need to calculate the magnification
Using formula of magnification
[tex]m=\dfrac{v}{u}[/tex]
[tex]m=\dfrac{22.21}{95}[/tex]
[tex]m=0.233[/tex]
The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.
