How would you determine the refractive index of the rectangular slab?

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Fix a sheet of white paper on the drawing board with the help of drawing pins at the four corners of the sheet.

Place a glass slab at the center of the paper and mark its boundary EFGH with a fine pencil.

Remove the glass slab. Draw any line AB making an angle of 40 degrees with the normal at the point B, the middle point of EF approximately.

Put the glass slab back in position on the boundary line. Fix two pins P1 and P2 vertically on the line AB approximately 5 cm apart such that one pin is close to the slab.

Look for the image of these pins in the slab from the opposite side GH and fix two pins P3 and P4 so that they are in the line with the image of P1 and P2 as seen through the slab and approximately 5 cm apart.

Join the holes of P3 and P4 to obtain the emergent ray. Draw a normal to GH at the point C. join BC to get the refracted ray.

g) Measure the angle of incidence and angle of refraction. Produce AB forward and draw a perpendicular from C on AB produced to meet it at P. Then the lateral displacement = CP.

Now you can calculate the r.i of the glass slab from the Snell’s law considering n1 = r.i. of Air = 1, μ=Sini/Sinr μ=Sini/Sinr  

Repeat the experiment with different angles of incidence 50 and 60 degrees, if you want.

HOPE THIS HELPS

HAVE A GOOD DAY  :)

ITZ RASPUTIN002