Find the refractive index of a liquid
Refractive index of a liquid
Aim
To find the refractive index of a liquid using convex lens and plane mirror.
Apparatus and Material Required
- Double convex lens (focal length nearly 20 cm),
- plane mirror (bigger in size than the aperture of the lens),
- laboratory stand fixed with a pin in clamp,
- metre scale,
- plumb line,
- water dropper.
Principle
A real and inverted image coincides with the object placed on the principal focus point of a convex lens when a plane mirror placed horizontally below the lens.
The rays from a pin AB placed on the principal focus F of a convex lens emerges out parallel to its axis. When these rays fall normally on a plane mirror placed horizontally below the convex lens, they retrace their path and form a real and inverted image A’B’ at the principal focal plane
of the lens. The size of image A’B’ is equal to the size of object pin AB and the tip of the pin gives the position of the second principal focus. Then \(f\) (OF) is the focal length of the convex lens (for
a thin lens) where O is the optical centre of the lens.
When the space between the lens and mirror is filled with a transparent liquid (say water) having refractive index, \(n_{wa}\), then the image formation of any object placed above the lens will be due to the combination of equiconvex lens (glass lens) and plano-concave lens(water). For this lens combination, focal length \(f’\) can be obtained by above method.
By the relation for thin lenses in contact, \[\frac{1}{f’} = \frac{1}{f}+\frac{1}{f_w}\] where \(\frac{1}{f_w}\) is the focal length of the plano-concave lens made of the transparent liquid
\(\frac{1}{f}\) is the focal length of the equiconvex lens
\(\frac{1}{f’}\) is the focal length of the combination of the lenses
\[\text{or, } \frac{1}{f_w} = \frac{1}{f’}-\frac{1}{f} \text{ ……… (i)}\]
from the lens maker’s formula for a plano-concave lens, \[\frac{1}{f_w} =( n_{wa}-1)\frac{1}{R}\] \[\frac{R}{f_w} =( n_{wa}-1)\] \[ n_{wa} = 1+ \frac{R}{f_w}\]
\[ n_{wa} = 1+ R\cdot \left( \frac{1}{f’}-\frac{1}{f}\right)\text{ ……from equation (i)}\]
where \(R\) is the radius of curvature of water lens i.e. convex lens.
Procedure
- Place the plane mirror on the base of a rigid laboratory stand keeping its reflecting surface upwards.
- Place the convex lens on the plane mirror.
- Fix a sharp-edged bright pin in the clamp and place it horizontally and above the lens. Adjust the position of the pin, using plumb, such that its tip B lies vertically above the optical centre of the convex lens.
- Shift the clamped pin gradually upward looking at the image and bring it to a height such that the tip B of the pin exactly coincides with the tip of its image B’.
- Note the distances of the pin from the upper and lower surfaces of the lens. Take its average. It is the measure of the distance OF. It is focal length \(f\) of the lens.
- Fill the space below lens with water using the dropper. Repeat the steps 4 and 5. It will give the focal length \(f’\) of the combination of two lenses.
- Repeat the experiments and note down the records.
Observations
Calculations
\(n_{wa} = 1+ R\cdot \left( \frac{1}{f’}-\frac{1}{f}\right)\)
\(n_{wa} \) = 0.833
Result
The refractive index of a given liquid (here water) with respect to air, \(n_{wa}\) is
Precautions
- Keep pin horizontal and its tip just above the optical center of the lens.
- Use thin lens only.
- Put water between the mirror and the lens gently to avoid air filling the space between them.
Sources of errors
- The plane mirror is not horizontal.
- Two surfaces of the convex lens do not have the same radius.
Conclusion
The experiment successfully determined the refractive index of a liquid using a convex lens and plane mirror by measuring the focal lengths of lens immersed in the air and the liquid.
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Find the refractive index of a liquid using convex lens and plane mirror.
