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Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refraction for refactive index of material of the prism. |
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Answer» Solution :(i) Let LIGHT ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and REFRACTION at the first face AB are `i_1` and `r_1` (ii) The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is `r_2` and `i_2` respectively. RS is the ray emerging from the second face. Angle `i_2` is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d. (iv) The two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M. The deviation d, at the surface AB is, ` angle RQM = d_1 = i_1 - r_1 "" ...(1)` The deviation d, at the surface AC is, ` angle QRM =d_2= i_2 - r_2 "" ..(2)` Total angle of deviation d produced is, ` d = d_1 + d_2`...(3) Substituting for `d_1` and `d_2` ` d = ( i_1 - r_1) + (i_2 -r_2)` After rearranging ` d = ( i_1- r_1) + (i_2 - r_2)`...(4) In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is `180^@` ` angle A+ angle QNR = 180^@`....(5) From the triangle `triagle QNR,` ` r_1 + r_2 + angle QNR = 180^@` COMPARING these two equations (4) and (5) we GET, ` r_1 + r_2 = A` Substituting this in equation (4) for angle of deviation, ` d = i_1 + i_2 - A "" ...(6)` At minimum deviation, `i_1 = i_2 = i` and ` r_1 = r_2 = r ` Now, theequation (6) becomes, `D = i_1 + i_2 - A = 2i - A " or " i = ((A+D))/(2)` The equation (5) becomes, ` r_1 + r_2 = A rArr 2r = A " or " r= A/2` Substituting i and r in snell's law ` n = (sin i )/(sin r)` ` n = (sin ((A+D)/(2)))/(sin (A/2))` |
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