#### Question

Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through a Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm?

#### Solution

Given:

`lambda = 10 "pm" = 100 xx 10^-12 "m"`

`D = 40 "cm" = 40 xx 10^-2 "m"`

`β = 0.1 "mm" = 0.1 xx 10^-3 "m"`

`β = (lambdaD)/d`

`d = (lambdaD)/β`

= `(100 xx 10^-12 xx 40 xx 10^-2)/(10^-3 xx 0.1)`

= `4 xx 10^-7 "m"`

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#### APPEARS IN

Solution Suppose a Monochromatic X-ray Beam of Wavelength 100 Pm is Sent Through a Young'S Double Slit and the Interference Pattern is Observed on a Photographic Plate Placed 40 Cm Away from the Slit. Concept: Electromagnetic Spectrum.