Unpolarized red light is incident on two identical, narrow vertical slits. The photograph at right shows the interference pattern that appears on a distant screen. a. Specify the quantity or quantities that are adding to zero at the interference minima. light waves from the two slits are adding to zero is not a sufficient answer.)b. A polarizer is placed directly in front of both slits, so that the light is vertically polarized before passing through the slits. It is observed that the intensity at each point on the screen decreases by a factor of two. How can your account for the decrease in intensity at the interference maxima?c. The polarizer is slowly rotated through 360 degree . As the polarizer is rotated, it is observed that the interference pattern does not change. What is the flaw in the reasoning? Explaind. Now imagine that one polarizer is placed in front of each slit: one polarizer with its direction of polarization vertical; the other, horizontal.Would there still be locations on the screen at which the intensity is zero? Explain why or why not.

Respuesta :

Electromagnetic waves make up light. Electric fields can be in any direction perpendicular to the propagation of unpolarized light rays. At each point, the intensity is halved.

a). Light is an electromagnetic wave. It consists of minimally intense, pulsing electric and magnetic fields that are directed in all directions perpendicular to the propagation diffraction pattern. Electric fields of light from two slits add up to zero, therefore they likewise sum to zero in terms of light intensity.

b). In light waves (unpolarized), the electric field can be in any direction perpendicular to its propagation. Polarization entails facing the electric field in all directions except one toward zero. While polarisation, for example, in the vertical direction, additional electric field components from other directions are added to the necessary (remaining) direction.

c). Electric fields at angle θ to vertical adds Ecosθ to vertical field

before polarisation: same E in all directions

I∝E²

∝[tex]\int\limits^\pi _0[/tex]E² dθ =E² π

After polarisation;

I∝ [tex]\int\limits^\pi _0[/tex] (E cosθ )²dθ =E² π/2

Hence, Intensity is halved at every point

d). Electric field cannot be stopped by slits since it is not a mechanical thing that can be.

e). The horizontal and vertical electric field vector addition cannot be zero. Therefore, there won't be any places with 0 intensity.

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