Both ends of a glass rod with an index of refraction 1.5 are ground in spherical shape. The radius of curvature at the left end is 5 cm and at the right end is 10 cm. The length of the rod between vertexes is 25 cm. The object for the surface at the left end is an arrow that lies 20 cm to the left of the vertex of this surface. The arrow is 1.4 mm tall and at right angles to the axis.
(i) What is the object distance and nature for the surface at the right end of the rod?
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Ans.
(i)
Sol. Image formed by refraction at first surface
– 
= 
⇒ v 1 = 30 cm to the right of first vertex.
This image serves as the object for the refraction at the second surface (right end of the rod). It is 30 – 25 = 5 cm right.
(ii)
Sol. Image formed by refraction at second surface
= 
⇒ v 2 = 
to the right of vertex at right hand of the
rod.
m = m 1 m 2 = 
× 
= 
overall magnification m is –ve means final image is
inverted with respect to the original object.
y ′ = my = – 
× 1.4 = – 1.2 mm
(iii)
Sol. Image formed after first refraction is 5 cm to
right of right end.
image formed by right end mirror
+ 
= – 
⇒ v 1 = – 2.5
i.e. 2.5 cm left of right end
Final image formed after second refraction
from left end is
– 
= 
⇒ 
⇒ v = 30 cm Left of left end of rod i.e. 10 cm to the
left of object
m = 
= + 1 erect
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