Suggested solution:
Let I 0 be the intensity of either of the interfering sources, the intensity I at a point P is where d is the phase difference at p between the two disturbances. (i) d = 60 0 ; I = I o cos 2 (30 0 ) (ii) d = 90 0 ; I = 4I o cos 2 (45 0 ) (iii) d = 120 0 ; I = 4 I o cos 2 (60 0..
Let I 0 be the intensity of either of the interfering sources, the intensity I at a point P is where d is the phase difference at p between the two disturbances. (i) d = 60 0 ; I = I o cos 2 (30 0 ) (ii) d = 90 0 ; I = 4I o cos 2 (45 0 ) (iii) d = 120 0 ; I = 4 I o cos 2 (60 0..Suggested solution:
Also d = 5 x 10 - 3 m, therefore..
Also d = 5 x 10 - 3 m, therefore..Suggested solution:
The two stars will just appear separate if the principal maximum due to one coincides with the first minimum of the other in diffraction pattern. In other words, the half angular width of diffraction maximum is equal to the angle subtended by the stars at the objective of telescope of d is the ape..
The two stars will just appear separate if the principal maximum due to one coincides with the first minimum of the other in diffraction pattern. In other words, the half angular width of diffraction maximum is equal to the angle subtended by the stars at the objective of telescope of d is the ape..Suggested solution:
According to Malus' law, the intensity I of transmitted light from an analyzer is I = I o cos 2 d Where s is the angle between the analyzer and the polarizer. Given I = I o / 2. Therefore, ..
According to Malus' law, the intensity I of transmitted light from an analyzer is I = I o cos 2 d Where s is the angle between the analyzer and the polarizer. Given I = I o / 2. Therefore, ..Suggested solution:
Since l is increasing, the star is receding from the earth. Let v be the speed of the star with respect to the earth: The apparent wavelength ' l ' is where c is the actual wave length. or v = 9 x 10 4 ..
Since l is increasing, the star is receding from the earth. Let v be the speed of the star with respect to the earth: The apparent wavelength ' l ' is where c is the actual wave length. or v = 9 x 10 4 ..Suggested solution:
u = -45 cm v = +90 cm, f = ? or The lens is clearly a converging lens. The negative sign indicates that the image is inverte..
u = -45 cm v = +90 cm, f = ? or The lens is clearly a converging lens. The negative sign indicates that the image is inverte..Suggested solution:
m = 1.55, R1 = R and R2 = -R, f = 20 cm or \ R = 40 x 0.55 cm = 22..
m = 1.55, R1 = R and R2 = -R, f = 20 cm or \ R = 40 x 0.55 cm = 22..Suggested solution:
emf of series combination of cells = 6 x 2 V = 12 V Total internal resistance of series combination = 6 x 0.015 W = 0.09 W Total resistance of circuit = (8.5 + 0.09) ohm = 8.59 ohm Terminal potential difference..
emf of series combination of cells = 6 x 2 V = 12 V Total internal resistance of series combination = 6 x 0.015 W = 0.09 W Total resistance of circuit = (8.5 + 0.09) ohm = 8.59 ohm Terminal potential difference..Suggested solution:
The valency of silver is one. So, silver is monoatomic. Thus, the chemical equivalent of silver is the same as the molar mass of silver. Since charge required to liberate one chemical equivalent of silver is 96500 C, therefore, the charge q required to liberate 0.5 chemical equivalent of silver is..
The valency of silver is one. So, silver is monoatomic. Thus, the chemical equivalent of silver is the same as the molar mass of silver. Since charge required to liberate one chemical equivalent of silver is 96500 C, therefore, the charge q required to liberate 0.5 chemical equivalent of silver is..Suggested solution:
Drift velocity, V d = ? Cross-sectional area, A = 5 x 10 - 6 m 2 Current, I = 10 A Number density of free electrons, n = 8 x 10 2 8 electrons/m 3 We know that o..
Drift velocity, V d = ? Cross-sectional area, A = 5 x 10 - 6 m 2 Current, I = 10 A Number density of free electrons, n = 8 x 10 2 8 electrons/m 3 We know that o..See what our Users say :
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