Both ends of a glass rod with index of refraction 1.60 are ground and polished t

Question

Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 25.0cm . The object for the surface at the left end is an arrow that lies 24.0cm to the left of the vertex of this surface. The arrow is 1.30mm tall and at right angles to the axis.

a) What is the object distance for the surface at the right end of the rod?

b) Is the object for this surface real or virtual? (Hint: If the light incident from the left onto a convex mirror does not diverge from an object point but instead converges toward a point at a (negative) distance s to the right of the mirror, this point is called a virtual object.)

c)What is the position of the final image?

d)Is the final image real or virtual?

e) Is it erect or inverted with respect to the original object?

f) What is the height of the final image?

Explanation / Answer

Solution:

a) Refractive index = n = 1.6

Radius of left surface = R1 = 6cm

Radius of curvature of right surface = R2 = 12 cm

Distance between the surfaces = d = 40 cm

Object distance to the right surface = d1o = 24 cm

Object to the right surface is the image of the left surface.

OBject distance to the right d2i = d - d1i (image distance of left surface )

1/d1o + n/d1i = (1.6 - 1) / R1

=> 1/24 + 1.6 /d1i = (1.6 - 1)/6

=> 1.6 / d1i = .6/6 - 1/24 = 0.058

=> d1i = 1.6/0.058 = 27.4 cm

Object distance of right surface = d2o = 40 - 27.4 = 12.6 cm

b) The object for this surface is real.

c) if dif = final image distance and d2o = object distance to the right surface

1.6 / (12.6) + 1/dif = (1-1.6) / -12

=> 0.12698 + 1/dif = 0.05

=> 1/dif = 0.05 - 0.12698 = -0.07698

=> dif = 1 / -0.07698 = - 12.99 cm = Final Image formed by system.

d) The final image is Virtual.

e) It is inverted with respect to the original object

f) h' = - ho / (n *d1o) = - (1.3) [ 27.4 / (1.6)*24) ] = -0.93 mm

Final magnitude = (-0.93)[(-1.6) (-12.99) /(1)(12.6) ] = = -(0.93)(1.6495) = - 1.53 mm

The final image formed is 1.53 mm high ;

So it is a virtual, Magnified , Inverted Image, with respect to the object on the left of the left surface.

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