With wavelength 560nm fringe width comes...

With wavelength 560nm fringe width comes out to be 72mm in a YDSE. Now if wavelength is changed so that fringe width is 81mm then new wavelength is equal to

490mm

630mm

800mm

700mm

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In a YDSE setup interference fringes are obtained by sodium light of wavelength 5890Å . On screen the fringes have an angular width 0.20^(@) . Now the wavelength of light is changed and it is found that fringe width increases by 10%, the new wavelength of incident light is :

In Young's double slit experiment, the fringe width with light of wavelength 6000 Ã… is 3 mm. The fringe width, when the wavelength of light is changed to 4000 Ã… is

Yellow light of wavelength 6000Å produces fringes of width 0.8 mm in YDSE. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500 Å and the separation between the slits is doubled ?

In YDSE wavelength of light used is 500 nm and slit width is 0.05 mm . then the angular fringe width will be.

In YDSE performed with wavelength lambda = 5890 Å the angular fringe width is 0.40^(@) . What is the angular fringe width if the entire set-up is immersed in water?

In Young's experiment , the fringe width of the fringes with light of wavelength 6000 Å is 2.0 mm . What will be the fringe width if the entire apparatus is immersed in a liquid of refractive index 1.33 ?

In YDSE setup slits are illuminated by a light of wavelength 4000Å and a light of unknown wavelength. It is observed that fourth dark fringe of known wavelength coincide with second bright fringe of unknown wavelength. Find the unknown wavelength.

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