Answer:
a) 17.15cm
b)0.687mm
c)erect
Explanation:
Given:
Diameter = 9.00 cm
[tex]n_{1}[/tex] for air =1
Index of refraction n₂= 1.61
Radius of curvature R= 4.50
Height of object h₀= 1.55 mm
Object distance u= 24.0 cm
(A)In order to calculate the image distance , we use formula for image of distance
[tex]\dfrac{n_{1}}{u}+\dfrac{n_{2}}{v}=\dfrac{n_{2}-n_{1}}{R}[/tex]
By plugging in all the required values
[tex]\dfrac{1}{24.0}+\dfrac{1.61}{v}=\dfrac{1.61-1}{4.50}[/tex]
[tex]\dfrac{1.61}{v}=\dfrac{1.61-1}{4.50}-\dfrac{1}{24.0}[/tex]
[tex]\dfrac{1.61 }{v}=\dfrac{169}{1800}[/tex]
[tex]v=\dfrac{1.61\times1800}{169}[/tex]
[tex]v=17.15\ cm[/tex]
(B). In order to calculate the height of the image , we'll use formula of magnification
[tex]m=\dfrac{h_{i}}{h_{o}}\dfrac{h_{i}}{h_{o}}=\dfrac{n_{1}d_{1}}{n_{2}d_{o}}[/tex]
By substituting all the required values, we get
[tex]\dfrac{h_{i}}{1.55}=\dfrac{1\times17.15}{1.61\times24.0}[/tex]
[tex]h_{i}=\dfrac{1\times17.15\times1.55}{1.61\times 24.0}\\h_{i}=0.687\ mm[/tex]
c) The image is erect as it is of (+)ve sign.
