A soap film has an index of refraction n = 1.50. The film is viewed in transmitt

Question

A soap film has an index of refraction n = 1.50. The film is viewed in transmittedlight. (a) At a spot where the film thickness is 950.0 nm, which wavelengths are missing in thetransmitted light?
(Enter your answers in order, least to greatest, from left toright. If there are more than 3 such visible wavelengths, enter the3 smallest.)
Enter anumber. 1 nm Enter anumber. 2 nm Enter anumber. 3 nm

(b) Which wavelengths are strongest in transmitted light?
(Enter your answers in order, least to greatest, from left toright. If there are more than 3 such visible wavelengths, enter the3 smallest.)
Enter anumber. 4 nm Enter anumber. 5 nm Enter anumber. 6 nm Enter anumber. Enter anumber. Enter anumber. Enter anumber. Enter anumber. Enter anumber.

Explanation / Answer

.      Refractive index is : n = 1.50 . (a) We know that : .       = 2 n t / (m + 1/2) .       = ( 2 * 1.50 * 950 nm) / (m + 0.5) .       = 2850 nm / ( m + 0.5 ) . For m = 3 ;    = 814.25 nm . For m = 4 ;    = 633.33 nm . For m = 5 ;    = 518.18 nm . For m = 6 ;    = 438.46 nm . We know that visible light range is from : 400 nm - 700 nm . Hence wave lengths are : 438.46 nm , 518.18 nm , 633.33 nm . (b) Strongest transmitted light is : .       = 2 n t / m = (2850 nm ) / m . For m = 3 ;    = 950 nm . For m = 4 ;    = 712.5 nm = 713 nm . For m = 5 ;    = 570 nm . For m = 6 ;    = 475 nm . For m = 7 ;    = 407 nm . Hence visible range is : 407 nm , 475 nm , 570 nm . Hope this helps u! For m = 3 ;    = 950 nm . For m = 4 ;    = 712.5 nm = 713 nm . For m = 5 ;    = 570 nm . For m = 6 ;    = 475 nm . For m = 7 ;    = 407 nm . Hence visible range is : 407 nm , 475 nm , 570 nm . Hope this helps u!

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