Respuesta :
Answer:
The position of the image and the height of the image are 19.68 cm and 0.828 mm.
Explanation:
Given that,
Diameter = 9.00 cm
Index of refraction n₂= 1.55
Radius of curvature R= 4.50
Height of object h₀= 1.50 mm
Object distance u= 23.0 cm
(A). We need to calculate the image distance
Using formula for image of distance
[tex]\dfrac{n_{1}}{u}+\dfrac{n_{2}}{v}=\dfrac{n_{2}-n_{1}}{R}[/tex]
Put the value into the formula
[tex]\dfrac{1}{23.0}+\dfrac{1.55}{v}=\dfrac{1.55-1}{4.50}[/tex]
[tex]\dfrac{1.55}{v}=\dfrac{1.55-1}{4.50}-\dfrac{1}{23.0}[/tex]
[tex]\dfrac{1.55}{v}=\dfrac{163}{2070}[/tex]
[tex]v=\dfrac{1.55\times2070}{163}[/tex]
[tex]v=19.68\ cm[/tex]
(B). We need to calculate the height of the image
Using formula of magnification
[tex]m=\dfrac{h_{i}}{h_{o}}[/tex]
[tex]\dfrac{h_{i}}{h_{o}}=\dfrac{n_{1}d_{1}}{n_{2}d_{o}}[/tex]
Put the value into the formula
[tex]\dfrac{h_{i}}{1.50}=\dfrac{1\times19.68}{1.55\times23.0}[/tex]
[tex]h_{i}=\dfrac{1\times19.68\times1.50}{1.55\times23.0}[/tex]
[tex]h_{i}=0.828\ mm[/tex]
Hence, The position of the image and the height of the image are 19.68 cm and 0.828 mm.
