A small object is embedded in a glass sphere ($\mu$ = 1.5) of radius 5.0 cm at a distance 1.5 cm left to the centre. Locate the image of the object as seen by an observer standing
a) to the left of the sphere and
b) to the right of the sphere
Answer
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Hint: The formula for the case of refraction from a concave surface has to be used here. As we are going to observe the sphere from left and right sides the object distance in both situations will be different.
Formula used:
For refraction from a spherical surface;
$\dfrac{\mu}{v} - \dfrac{1}{u} = \dfrac{(\mu - 1)}{R}$
where symbols have their usual meaning.
Complete answer:
We are given the radius of the glass sphere R = 5 cm. The refractive index of the glass $\mu$ = 1.5. Keeping the object at a distance 1.5 cm to the left of the centre, for the two cases to be considered i.e., observer to the left of the sphere and observer to the right, we find the object distance to be; (a) u = - (5 - 1.5)= -3.5 cm and (b) u = -(5+1.5) cm = - 6.5 cm respectively. For our convenience, we shift the observer to the same side in both cases and draw the diagram as:
As we go from glass to sphere, we write:
$\mu = \dfrac{1}{1.5}=0.66$.
Now, we obtain the image distance as;
(a) for the observer to the left,
$\dfrac{0.66}{v} - \dfrac{1}{-3.5 cm} = \dfrac{(0.66 - 1)}{-5 cm}$
$\implies \dfrac{1.5}{v} = - \dfrac{1}{3.5} + \dfrac{0.34}{5}$
$\implies \dfrac{0.66}{v} = \dfrac{-5 + 0.34 \times 3.5}{3.5 \times 5} $
$\implies v = - \dfrac{3.5 \times 5 \times 0.66}{3.81} = - 3.29 cm$
Therefore, as we keep the object at a distance of 3.5 cm from the concave glass surface, it forms an image inside the glass surface at a distance of 3.29 cm from the surface.
(b) Similarly, for the observer to the right, we get:
$\dfrac{0.66}{v} - \dfrac{1}{-6.5 cm} = \dfrac{(0.66 - 1)}{-5 cm}$
$\implies \dfrac{1.5}{v} = - \dfrac{1}{6.5} + \dfrac{0.34}{5}$
$\implies \dfrac{0.66}{v} = \dfrac{-5 + 0.34 \times 6.5}{6.5 \times 5} $
$\implies v = - \dfrac{6.5 \times 5 \times 0.66}{2.79} = - 7.69 cm$
Therefore, as we keep the object at a distance of 3.5 cm from the concave glass surface, it forms an image inside the glass surface at a distance of 7.69 cm from the surface.
Note:
For the second case too, we considered only refraction from a concave surface and nothing else. For the case of a thin lens, we consider the refraction from the second surface too. Remember that our object is placed inside the sphere and the incident light from that object will hit the concave surface causing refractive. Therefore we measure our object and image distance from the concave glass surface instead of from the centre of the sphere.
Formula used:
For refraction from a spherical surface;
$\dfrac{\mu}{v} - \dfrac{1}{u} = \dfrac{(\mu - 1)}{R}$
where symbols have their usual meaning.
Complete answer:
We are given the radius of the glass sphere R = 5 cm. The refractive index of the glass $\mu$ = 1.5. Keeping the object at a distance 1.5 cm to the left of the centre, for the two cases to be considered i.e., observer to the left of the sphere and observer to the right, we find the object distance to be; (a) u = - (5 - 1.5)= -3.5 cm and (b) u = -(5+1.5) cm = - 6.5 cm respectively. For our convenience, we shift the observer to the same side in both cases and draw the diagram as:
As we go from glass to sphere, we write:
$\mu = \dfrac{1}{1.5}=0.66$.
Now, we obtain the image distance as;
(a) for the observer to the left,
$\dfrac{0.66}{v} - \dfrac{1}{-3.5 cm} = \dfrac{(0.66 - 1)}{-5 cm}$
$\implies \dfrac{1.5}{v} = - \dfrac{1}{3.5} + \dfrac{0.34}{5}$
$\implies \dfrac{0.66}{v} = \dfrac{-5 + 0.34 \times 3.5}{3.5 \times 5} $
$\implies v = - \dfrac{3.5 \times 5 \times 0.66}{3.81} = - 3.29 cm$
Therefore, as we keep the object at a distance of 3.5 cm from the concave glass surface, it forms an image inside the glass surface at a distance of 3.29 cm from the surface.
(b) Similarly, for the observer to the right, we get:
$\dfrac{0.66}{v} - \dfrac{1}{-6.5 cm} = \dfrac{(0.66 - 1)}{-5 cm}$
$\implies \dfrac{1.5}{v} = - \dfrac{1}{6.5} + \dfrac{0.34}{5}$
$\implies \dfrac{0.66}{v} = \dfrac{-5 + 0.34 \times 6.5}{6.5 \times 5} $
$\implies v = - \dfrac{6.5 \times 5 \times 0.66}{2.79} = - 7.69 cm$
Therefore, as we keep the object at a distance of 3.5 cm from the concave glass surface, it forms an image inside the glass surface at a distance of 7.69 cm from the surface.
Note:
For the second case too, we considered only refraction from a concave surface and nothing else. For the case of a thin lens, we consider the refraction from the second surface too. Remember that our object is placed inside the sphere and the incident light from that object will hit the concave surface causing refractive. Therefore we measure our object and image distance from the concave glass surface instead of from the centre of the sphere.
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