Objective:verify the laws of reflection using a glass a slab and find the refractive index of the refractive index of the material.
Answers
Answer:
The incident ray, the refracted ray and the normal to the surface at the point of incidence all lie in one plane. For any two given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
sinrsini=μ
Where μis the refractive index of the second medium with respect to the first medium.
Place a rectangular glass slab on the white sheet of paper fixed on a drawing board.
Trace the boundary ABCD of the glass slab.
Remove the glass slab and draw a normal N1N2 at O.
Draw a straight line IO inclined at an angle say 300 with the normal. IO is the incident ray.
Fix two pins P and Q on the incident ray IO.
Place the glass slab within its boundary ABCD.
Looking from the other side of the glass slab fix two other pins R and S such that P, Q, R and S appear to lie on the same straight line.
Remove the glass slab and the pins. Mark the pin points P, Q, R and S.
Join the pins R and S and produce the line on both sides. The ray O'E is the emergent ray.
Join OO'. It is the refracted ray.
The incident ray, the refracted ray and the normal are all lying in the same plane.
This proves the first law of refraction.
Let us now prove the second law of refraction
With O as center, draw a circle of a convenient radius 'R' in such a way that it cuts the incident and the refracted rays at F and G respectively.
From F and G draw perpendiculars to the normal N1N2.
Triangle FHO and triangle GKO are right-angled triangles.
sini=OFFH
sinr=OGGK
μ=sinrsini=OFFH×GKOG
But OG=OF=R
μ=OGFH×GKOG
μ=GKFH
Measure the length of FH and GK and record them in the observation table.
Repeat the experiment for different values of angle of incidence.
Find the value of GKFH for different values of i.
In each case it is found that the ratio GKFH is the same, that is, the ratio of GKFH is a constant. This verifies law of refraction.
Explanation:
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