Angle of incidence can be changed in steps of 5° from 0° to 75°
Click in the left half of the circle decreases the angle of incidence by 5°
Click in the right half of the circle increases the angle of incidence by 5°
Please increase the angle of incidence to the maximum value for a graph to appear.
Up and down arrows on either side of μ label change the value of refractive index from 1.3 to 1.8
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Figure shows the path of a ray of light through an equiangular prism. The incident ray makes an angle \( i_1 \) with the normal to face AB. The refracted ray bends towards the normal making an angle \( r_1 \) with it. It is then incident at \(r_2 \) on the face AC and and emerges at \( i_2 \).
As the ray refracts at the face AB it undergoes a deivation of \( i_1-r_1 \) towards the base BC and when it refracts the second time it deviates by \( i_2-r_2 \) again towards the base. The ray suffers a total deviation of \(d = (i_1-r_1) + (i_2-r_2) \).
It can be shown that the sum of \(r_1 + r_2 \) is equal to the angle of the prism A. So the deviation \(d = i_1 +i_2 - A \).
Graph shows the variation of angle of deviation with the angle of incidence. As the angle of incidence increases, the angle of deviation decreases, reaches a minimum value \( (d_m) \) and then increases. \( i_1 \) and \( i_2 \) are the interchangeable angles of incidence and emergence. When the deviation is minimum these two angles \( i_1 \) and \( i_2 \) are equal ( \(i \) say )
So \( d_m = i+i-A \) or \( i = \, (A+d_m)/2 \)
Also the two corresponding angles of refraction are equal. \( r_1 = r_2 = r \) (say). So \( A = r+r = 2r \) or \( r = A/2 \)