Question

The left end of a long glass rod 9.00 cm in diameter, with an index of refraction 1.61, is ground and polished to a convex hemispherical surface with a radius of 4.50 cm. An object in the form of an arrow 1.55 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface.(a) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.(b) Find the height of the image formed by paraxial rays incident on the convex surface.(c) Is the image erect or inverted?

Answer 1

a) 17.15cm

b)0.687mm

c)erect

Step-by-step explanation:

Given:

Diameter = 9.00 cm

`n_(1)` 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

`(n_(1))/(u)+(n_(2))/(v)=(n_(2)-n_(1))/(R)`

By plugging in all the required values

`(1)/(24.0)+(1.61)/(v)=(1.61-1)/(4.50)`

`(1.61)/(v)=(1.61-1)/(4.50)-(1)/(24.0)`

`(1.61 )/(v)=(169)/(1800)`

`v=(1.61*1800)/(169)`

`v=17.15\ cm`

(B). In order to calculate the height of the image , we'll use formula of magnification

`m=(h_(i))/(h_(o))(h_(i))/(h_(o))=(n_(1)d_(1))/(n_(2)d_(o))`

By substituting all the required values, we get

`(h_(i))/(1.55)=(1*17.15)/(1.61*24.0)`

`h_(i)=(1*17.15*1.55)/(1.61* 24.0)\\h_(i)=0.687\ mm`

c) The image is erect as it is of (+)ve sign.