Question

2.) The image of a red (620 nm) text display is to be mirrored into the path of light of a projector. This is done with a glass plate in 45 degrees covered with a thin layer with n=1.80. How thick should this layer be in order to get maximum reflectance for red but minimal for blue through green?

Answer 1

phase shift Hw two reflections 4P 2T and 1os Ot il. here TT - odd nomber of reflecfions From snelf law:- =nsinot no = sin8; no sin @p 12H => Sino o ho sin O; sin sin I 45) - sin 1.80 $3.13 for constructive Enter ferenie A9 i multiple of 2l. 2T end los ot tl (2m-1) 4h co or

irtructive nlerfernck &nd cos @f tT Amin Cam -1) dr -> dmin m=1,2, 3 Ca0)-1) tor 620 nM d min = X620 310 nm 4-1 X 620 nm Amin = n m 3 x 620 = 465 6-| Y 620 Amin - 516.66 m for m-1 reflets both blue& green Maximum blue for m= 3 dees n't refled three. m= Q is the best among So (v620 (a m - Dir

(go) -1) dr 4n1os or 1620nm 4nlosot nim X 620nm. 4x1,80 Y (s (23,13) 280.91 nm 1.5 please solution Rate Thank U. my