Explanation:
Unequal width of slits will cause unequal intensity of lights entering from both slits.
As a result, during interference complete cancelling of light intensity will not take place at regions of otherwise dark fringe.
As the value of $\beta$ does not depend on intensity of light, there will be no shifting of fringes as well as no change in fringe width.
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What is increased in step-down transformer
|
(a) Voltage |
(b) Current |
(c) Power |
(d) Current density |
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern
|
(a) The intensities of both the maxima and the minima increase |
|
(b) The intensity of maxima increases and the minima has zero intensity |
|
(c) The intensity of maxima decreases and that of the minima increases |
|
(d) The intensity of maxima decreases and the minima has zero intensity |
The ratio of secondary to primary turns is 9 : 4. If power input is P, what will be the ratio of power output (neglect all losses) to power input
|
(a) 4 : 9 |
(b) 9 : 4 |
(c) 5 : 4 |
(d) 1 : 1 |
When a slow neutron goes sufficiently close to a 
nucleus, then the process that takes place is
|
(a) Fission of |
(b) Fusion of neutron |
(c) Fusion of |
(d) First (a) then(b) |
A current I is passing through a wire having two sections P and Q of uniform diameters d and d/2 respectively. If the mean drift velocity of electrons in sections P and Q is denoted by vP and vQ respectively, then
|
(a) vP = vQ |
(b) vP = |
(c) vP = |
(d) vP = 2 vQ |
