Question

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.​

Answer 1

Answer:

Distance between the slits, d=0.28×10

−3

m

Distance between the slits and the screen, D=1.4m

Distance between the central fringe and the fourth (n=4) fringe,

u=1.2×10

−2

m

In case of a constructive interference, we have the relation for the distance between the two fringes as:

u=nλD/d

⇒λ=ud/nD=6×10

−7

m=600nm

IF UH FOUND IT HELPFUL THEN PLEASE MARK AS BRAINLIEST.

Answer 2

AnSwer :

Distance between the slits, d = 0.28mm = 0.28 × 10¯³m.

Distance between the slits and the screen,D = 1.4m.

Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2cm = 1.2 × 10¯²m.

In case of constructive interference we have the revolution for the distance between the two fringes as :

${ \boxed{ \bf{u = n\lambda \dfrac{D}{d}}}}$

where,

n = order of fringes = 4

$\sf \lambda$ = wavelength of light used

${ \leadsto{ \bf{\lambda = \dfrac{ud}{nD}}}}$

${ \leadsto{ \bf{\lambda = \dfrac{1.2 \times {10}^{ - 2} \times 2.8 \times {10}^{ - 3} }{4 \times 1.4}}}}$

${ \leadsto{ \bf{\lambda = 6 \times {10}^{ - 7} }}}$

${ \leadsto{ \bf{\lambda = 600nm}}}$

thus, the wavelength of light is 600nm