Air bubble inside a glass sphere problem

[Amith2006]

Sir,
Please help me with this problem.
# A small air bubble in a sphere of 8 cm diameter of a substance having refractive index 1.4 appears to be 2 cm from the surface when looking along the diameter. Find the true position of the bubble.
I solved it in the following way:
Here n1 = 1, n2 = 1.4, v = -2 cm(Virtual image), R = + 4 cm
n2/v + n1/u = (n2 – n1)/R
1.4/-2 + 1/u = 0.4/4
1/u = 0.1 + 0.7 = 0.8
u = 1/0.8
u = 1.25 cm
But the book answer is 2.33 cm. But I get this answer if I do it in a different way.
Here n1 = 1, n2 = 1.4, v = -2 cm(Virtual image), R = + 4 cm
n1/v + n2/u = (n1 – n2)/R
1/-2 + 1.4/u = (1.4 – 1)/4
1.4/u = 0.1 + 0.5 = 0.6
u = 1.4/0.6
u = 2.33 cm

 

[Doc Al]

I solved it in the following way:
Here n1 = 1, n2 = 1.4, v = -2 cm(Virtual image), R = + 4 cm
n2/v + n1/u = (n2 – n1)/R
1.4/-2 + 1/u = 0.4/4
1/u = 0.1 + 0.7 = 0.8
u = 1/0.8
u = 1.25 cm

I don't understand this solution; seems like you mixed up n1 and n2.

But the book answer is 2.33 cm. But I get this answer if I do it in a different way.
Here n1 = 1, n2 = 1.4, v = -2 cm(Virtual image), R = + 4 cm
n1/v + n2/u = (n1 – n2)/R
1/-2 + 1.4/u = (1.4 – 1)/4
1.4/u = 0.1 + 0.5 = 0.6
u = 1.4/0.6
u = 2.33 cm

I'd say that you made two errors in this one that counterbalance each other. (Unless I'm misreading what you've done.)

Using the usual (Halliday & Resnick) sign convention, the equation you need is:
n1/u + n2/v = (n2 - n1)/R
where u = object distance; v = image distance

Since the light from the object goes from inside the sphere to outside, n1 = 1.4 & n2 = 1. Also, since the surface as the light hits it is concave, R = -4 cm. Given is that v = - 2 cm (a virtual image). Solve for u, the object distance.