The r.m.s value of Potential due to superposition of given two alternating potentials `E_(1) = E_(0) sin omega t` and `E_(2) = E_(0) cos omega t` will be

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

Three light waves combine at a certain point where thcir electric field components are E_(1) = E_(0) sin omega t , E_(2)= E_(0) sin(omega t +60^(@)) , E_(3)= E_(0) sin(omega t -30^(@)) . Find their resultant component E(t) at that point.

Two coherent sources emit light waves which superimpose at a point where these can be expressed as E_(1) = E_(0) sin(omega t + pi//4) E_(2) = 2E_(0) sin(omega t - pi//4) Here, E_(1) and E_(2) are the electric field strenghts of the two waves at the given point. If I is the intensity of wave expressed by field strenght E_(1) , find the resultant intensity

Two coherent sources emit light waves which superimpose at a point where these can be expressed as E_(1) = E_(0) sin(omega t + pi//4) E_(2) = 2E_(0) sin(omega t - pi//4) Here, E_(1) and E_(2) are the electric field strenghts of the two waves at the given point. If I is the intensity of wave expressed by field strenght E_(1) , find the resultant intensity

Two waves at a point are represented by E_(1)= E_(0) sin omega t and E_(2)=E_(0) sin ( omega + phi) . There will eb destructive interference at this point is

In the given figure, the instantaneous value of alternating e.m.f. is e = 14.14 sin omega t . The reading of voltmeter in volt will be