Consider the arrangement shown in figure. By some mechanism, the separation between the slits $$S_3$$ and $$S_4$$ can be changed. The intensity is measured at the point $$P$$ which is at the common perpendicular bisector of $$S_1 S_2$$ and $$S_3 S_4$$. When $$z=\dfrac{D\lambda}{2d}$$, the intensity measured at $$P$$ is $$I$$. Find this intensity when $$z$$ is equal to (a) $$\dfrac{D\lambda}{d}$$ (b) $$\dfrac{3D\lambda}{2d}$$ and $$\dfrac{2D\lambda}{d}$$
In a Young's double slit experiment, the path different, at a certain point on the screen, between two interfering waves is $$\dfrac{1}{8}th$$ of wavelength. The ratio of the intensity at this point to that at the centre of a brigth fringe is close to :
If a Young's double slit experiment with light of wavelength $$\lambda$$ the separation of slits is d and distance of screen is $$D$$ such that $$D > > d > > \lambda$$. If the Fringe width is $$\beta$$, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is
The intensity at the maximum in a Young's double slit experiment is $${I}_{0}$$. Distance between two slits is $$d = 5 \lambda$$, where $$\lambda$$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $$D=10\ d$$?
While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $$\left( { S }_{ 1 },{ S }_{ 2 } \right) $$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the $$x-z$$ plane (for $$z > 0$$) at a distance $$D=3 m$$, from the mid-point of $$ { S }_{ 1 },{ S }_{ 2 }$$, as shown schematically in the figure. The distance between the sources $$d=0.6003 mm$$. The origin O is at the intersection of the screen and the line joining $$\displaystyle { S }_{ 1 },{ S }_{ 2 }$$. Which of the following is(are) true of the intensity pattern on the screen?
shown three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $$B{P_0} - A{P_0} = \lambda /3$$ and $$d > > \lambda $$ . show that in this case $$d = \sqrt {2\lambda D/3} $$ .(b) show that the intensity at $${P_0}$$ is three times the intensity due to any of the three slits individually.
The slits in a Young's double slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringes is $$I_0$$. If one of the slits closed, the intensity at this point will be
In Young's double-slit experiment, the slits are horizontal. The intensity at a Point P is $$\dfrac{3}{4}I_0$$, 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 $$2 \lambda$$)
(a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if $$ \lambda=550 \mathrm{nm}, d=0.150 \mathrm{mm} $$ and $$ a=30.0 \mu \mathrm{m} ? $$ (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?
In Young's double slit interference experiment the slit separation is made $$3$$ fold. The fringe width becomes
In a Young's double slit experimental arrangement shown here, if a mica sheet of thickness $$t$$ and refractive index $$\mu$$ is placed in front of the slit $$S_{1},$$ then the path difference $$(S_{1}P-S_{2}P)$$