Answer:
v = 6.315 cm
Explanation:
given,
R₁ = 4 cm = 0.04 m
R₂ = 15 cm = 0.15 m
n =1.5
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})[/tex]
[tex]\dfrac{1}{v}-\dfrac{1}{-1}=(1.5-1)(\dfrac{1}{0.04}+\dfrac{1}{0.15})[/tex]
[tex]\dfrac{1}{v}+1 = 0.5 \times 31.66[/tex]
[tex]\dfrac{1}{v} = 15.833[/tex]
v = 0.06315 m
v = 6.315 cm
hence, the distance of the image from the back surface is v = 6.315 cm
