A thin film of specific material can be used to decrease the intensity of reflected light.There is destructive interference fo waves from uppe rand lower surface of the fil,m.These films are called non-reflecting coatings.The process of coating the lens of or surface with non-reflectig film is called bloomingas shown in fig. Magnesium fluoride`(MgF_(2))` is genertally used as anti-reflectioncoating.If refrative index of `MgF_(2)` is `1.35` then minimum thickness of film required is `(Take lambda=550nm)`A. `112.4mm`B. `78.2mm`C. `99.6mm`D. None

A thin film of specific material can be used to decrease the intensity

1.	A thin film of specific material can be used to decrease the intensity of reflected light.There is destructive interference fo waves from uppe rand lower surface of the fil,m.These films are called non-reflecting coatings.The process of coating the lens of or surface with non-reflectig film is called bloomingas shown in fig. Magnesium fluoride`(MgF_(2))` is genertally used as anti-reflectioncoating.If refrative index of `MgF_(2)` is `1.35` then minimum thickness of film required is `(Take lambda=550nm)`A. `112.4mm`B. `78.2mm`C. `99.6mm`D. None
Answer» Correct Answer - C

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Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength `6000Å`. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will beA. `3500 Å`B. `4200 Å`C. `4700 Å`D. `6000 Å`
Find the half angular width of the central bright maximum in the fraunhofer diffraction pattern of a slit of width `12xx10^(-5)cm` when the slit is illuminated by monochromatic light of wavelength 6000 Å.
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Light wavelength 6000 Å is incident normally on a slit of width `24xx10^(-5)` cm. find out the angular position of seconds minimum from central maximum?

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