A plane monochromatic light wave with intensity `I_(0)` falls normally on an opaque screen with a round aperture. What is the intensity of light `I` behind the screen at the point for which the aperture
(a) is equal to the first Fresnel zone, to the internal half of the first zone,
(b) was made equal to the first Fresnel zone and then half of it was closed (along the diameter) ?

A plane monochromatic light wave with intensity I_(0) falls normally on an opaque disc closing the first Fresnel zone for the observation point P . What did the intensity of light I at the point P become equal to after (a) half of the disc (along the diameter) was removed, (b) half of the external half of the first Fresnal zone removed (along the diameter)?

A plane light wave with wavelength lambda and intensity I_(0) falls normally on a large glass plate whose opposite side serves as an opaque screen with a round aperture equal to the first Fresnel zone for the observation point P . In the middle of the aperture there is a round recess equal to half the Fresnel zone. What must the depth h of that recess be for the intensity of light at the point P to be the highest? What is this intensity equal to?

A plane monochromatic light wave with intensity I_(0) falls normally on the surfaces of the opaque screens shown in Fig. Find the intensity of light I at a point P (a) located behind the corner ponts of screens 1-3 and behind the edge of half-plane 4 , (b) for which the rounded-off edge of screens 5-8 coincides with the boundary of the first formula describing the result obtained for screens 1-4 , the same, for screens 5-8 .

If the distance between a point source and screen is doubled, then intensity of light on the screen will become

In a biprism experiment, calculate the intensity of lights on the screen at a point 5beta//6 away from the central fringe in terms of the intensity of the central fringe, which is I_(0) . Here beta is the fringe width.