IIT-JEE 2012 · Paper 1 · Physics Q6

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PhysicsSingle correct+3 / -1Question 6 of 60

Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_G$ , $\beta_R$ and $\beta_B$ , respectively. Then,

\beta_G>\beta_B>\beta_R

\beta_B>\beta_G>\beta_R

\beta_R>\beta_B>\beta_G

\beta_R>\beta_G>\beta_B

Answer and worked solution

Official answer: D

Fringe width $\beta=\dfrac{\lambda D}d\propto\lambda$ , and $\lambda_R>\lambda_G>\lambda_B$ . So $\beta_R>\beta_G>\beta_B$ .

(D)

Topics:Interference and Young's double slit

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IIT-JEE 2012 · Paper 1, Physics question 6