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Explain refraction of light from a denser toa rear medium, using the concept of wavefronts. |
Answer» Solution : where, `BC=v_1 tau,AD=v_2 tau,v_2 gt v_1,tau=` time taken to travel optical path by light. AB - incident wave front DC = refracted wave front. i = angle of incidence in medium (1) For `I ge i_e` , total internal reflection takes PLACE at the INTERFACE PQ. For `i LT i_e`, there will be refraction of light fron denser to rarer medium From the right angled triangle ABC `sin i=(BC)/(AC)=(v_1 tau)/(AC)""...(1)` and from the right angled triangle ADC, `sin r =(AD)/(AC)=(v_2 tau)/(AC)""...(2)` `(1) div (2)` gives `(sin i)/(sinr) =(v_1)/(v_2)""...(3)` Since, `v_1 prop (1)/(n_1) and v_2 prop (1)/(n_2)` equation (3) may be written as `(sin i)/(sin r)=(n_2)/(n_1)` where , `n_2 lt n_1` THEREFORE n_1 sin i=n_2 sin r`. |
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