One end of a long glass rod (n = 1.50) is formed into a convex surface with a ra

Question

One end of a long glass rod (n = 1.50) is formed into a convex surface with a radius of curvature of magnitude 6.50 cm. An object is located in air along the axis of the rod.

(a) Find the image position corresponding to object distance of 21.0 cm from the convex end of the rod. __________cm

(b) Find the image position corresponding to object distance of 11.0 cm from the convex end of the rod. ___________cm

(c) Find the image position corresponding to object distance of 3.3 cm from the convex end of the rod. ____________cm

Explanation / Answer

The expression for imaging surface is convex

n1/o + n2/i = (n2-n1)/R

Where

n1 n2 They are the refractive index of the two media, air and rod

o is the object distance

i is the image distance

R is the radius of curvature of the convex surface

Data

n2 =1.5

R = 6.50 cm

Part a)

o = 21 cm

n2/i = (n2-n1)/R – n1/o

n2/i = ( 1.50 – 1)/ 6.50 - 1/21

n2/i = 0.0769 - 0.0476 = 0.0293

i = n2/0.0293

i = 1.5/0.0293

i = 51.19 cm

real image

Part b)

o= 11 cm

n2/i = ( 1.50 – 1)/ 6.50 - 1/11

n2/i= 0.0769 - 0.0909

n2/i = - 0.014

i = - 1.50/0.014

i = - 107.14 cm

virtual image

Part c)

o = 3.3 cm

n2/i = ( 1.50 – 1)/ 6.50 - 1/ 3.3

n2/i= 0.0769 - 0.3030

n2/i = - 0.2261

i = - 1.50/0.2261

i = - 6.634 cm

virtual image

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