In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is 0.1° . If wavelength of light used is 6000 mathringA then spacing between the slits is:

Question

In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is 0.1° . If wavelength of light used is 6000 mathringA then spacing between the slits is: (A)  3.44× 10-4m  (B)  4.33× 10-4m  (C)  5.44× 10-4m  (D)  6.33× 10-4m

Tardigrade Admin · Accepted Answer

Angular fringe width α=0.1° =(0.1 × π/180) =1.74 × 10-3 rad λ =6000 mathringA =6000 × 10-10 m =6 × 10-7 m The angular fringe width is α-(λ/d) (where d is the separation between the slits) d =(λ/α) =(6 × 10-7/1.746 × 10-3) =3.44 × 10-4 m

Q. In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is $0.1^{\circ}$ . If wavelength of light used is $6000 \overset{˚}{A}$ then spacing between the slits is:

$3.44 \times 10^{- 4} m$

$4.33 \times 10^{- 4} m$

$5.44 \times 10^{- 4} m$

$6.33 \times 10^{- 4} m$

Q. In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is 0.1∘ . If wavelength of light used is 6000A˚ then spacing between the slits is:

3.44×10−4m

4.33×10−4m

5.44×10−4m

6.33×10−4m

Angular fringe width α=0.1∘ =1800.1×π​ =1.74×10−3rad λ=6000A˚ =6000×10−10m =6×10−7m The angular fringe width is α−dλ​ (where d is the separation between the slits) d=αλ​ =1.746×10−36×10−7​ =3.44×10−4m

Q. In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is $0.1^{\circ}$ . If wavelength of light used is $6000 \overset{˚}{A}$ then spacing between the slits is:

$3.44 \times 10^{- 4} m$

$4.33 \times 10^{- 4} m$

$5.44 \times 10^{- 4} m$

$6.33 \times 10^{- 4} m$