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27)
find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D
29)
For a stationary wave in P there should be a node at the closed end and an antinode at the open end.For a wave to be formed in Q there should be an antinode at both ends.
length of P=35 cm
wavelength of given wave = 20 cm
so 35/20 = 1.75 lambda can fit in P. if you draw a wave 1.75 lambda long starting with a node you will find that it ends with an antinode so this means that a stationary wave would be formed in P
for Q 50/20 = 2.5 lambda would fit in it.
start by drawing an antinode at one open end and continue until you draw 2.5 lambda .u will find that an antinode also forms at the other end.This satisfies the condition for a stationary wave .hence you will get a stationary wave in Q as well