Home
Class 12
PHYSICS
In the figure shown S(1)O-S(2)O=S(3)O-S(...

In the figure shown `S_(1)O-S_(2)O=S_(3)O-S_(2)O=(lambda)/(4)`, Intensity at O due to any one of the slits is `I_(0)`. What is the intensity due to all the three coherent sources `S_(1), S_(2) and S_(3)` ?

Text Solution

Verified by Experts

Phase difference corresponding to the given path difference.
`triangle phi= (2pi)/(lambda)* triangle x= (2pi)/(lambda)xx(lambda)/(4)= (pi)/(2)`
Let `A_(0)` be the amplitude due to either of the slits.
`A_(R )= sqrt(5)A_(0), I_(R )" as "I propto A^(2)`.
.
Promotional Banner

Topper's Solved these Questions

  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA|54 Videos

  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA (DIFFRACTION)|33 Videos

  • WAVES

    AAKASH SERIES|Exercise EXERCISE-III (Doppler effect :)|15 Videos

Similar Questions

Explore conceptually related problems

S_(8)+NaOH to Na_(2)S+Na_(2)S_(2)O_(3)

S_(8)+NaOH to Na_(2)S+Na_(2)S_(2)O_(3)

A monochromatic light source of wavelength lamda is placed at S. three slits S_(1),S_(2) and S_(3) are equidistant from the source S and the point P on the screen S_(1)P-S_(2)P=lamda//6 and S_(1)P-S_(3)P=2lamda//3 . if I be the intensity at P when only one slit is open the intensity at P when all the three slits are open is-

In the reaction, I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(-)+S_(4)O_(6)^(2-) .

In the arrangement shown in Fig., slits S_(1) and S_(4) are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of S_(1) S_(2) and S_(3) S_(4) . When Z = (lambda D)/(2d) , the intensity measured at O is I_(0) . The intensity at O When Z = (2 lambda D)/(d) is

I_(2)+S_(2)O_(3)^(2-) to I^(-)+S_(4)O_(6)^(2-)

I_(2)+S_(2)O_(3)^(2-) to I^(-)+S_(4)O_(6)^(2-)

In the arrangement shown in Fig., slits S_(1) and S_(4) are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of S_(1) S_(2) and S_(3) S_(4) . If a hole is made at O' on AO' O and the slit S_(4) is closed, then the ratio of the maximum to minimum observed on screen at O , if O' S_(3) is equal to (lambda D)/(4d) , is

In Young's double slit experiment, the slits are horizontal. The intensity at a point P as shown in figure is (3)/(4) I_(0) ,where I_(0) is the maximum intensity. Then the value of theta is, (Given the distance between the two slits S_(1) and S_(2) is 2 lambda )

Na_(2)S_(2)O_(3) is prepared by :