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**JEE main Previous Year Topic Wise Questions of Physics with Solutions**

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**Previous Years JEE Advanced Questions**

Q.

**Column I**shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits*$S_{1}$*and*$S_{2}$*. In each of these cases*$S_{1} P_{0}$*=*$S_{2} P_{0}$*,*$S_{1} P_{1}$*–*$S_{2} P_{1}$*= $\lambda / 4$ and*$S_{1} P_{2}$*–*$S_{2} P_{2}=\lambda / 3$*, where $\lambda$ is the wavelength of the light used. In the cases*B*,*C*and*D*, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point*P*on the screen from the two slits is denoted by $\delta$ (*P*) and the intensity by*I*(*P*). Match each situation given in**Column I**with the statement(s) in**Column II**valid for that situation.**[IIT-JEE-2009]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**((A) $p, s ;(B) q ;(C) t ;(D) r, s, t$)(A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$$\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$$\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$$\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$$\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$$\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$$(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$$\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$$\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$$8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$$8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$$\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$$=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$$8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$

Q. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_{G}, \beta_{R}$ and $\beta_{B},$ respectively. Then(A) $\beta_{G}>\beta_{B}>\beta_{R}$(B) $\beta_{B}>\beta_{G}>\beta_{R}$(C) $\beta_{R}>\beta_{B}>\beta_{G}$(D) $\beta_{R}>\beta_{G}>\beta_{B}$

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**Sol.**(D)$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$$\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$

Q. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is :-(A) $(2 n+1) \frac{\lambda}{2}$(B) $(2 n+1) \frac{\lambda}{4}$(C) $(2 n+1) \frac{\lambda}{8}$$(D)(2 n+1) \frac{\lambda}{16}$

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**Sol.**(B)$\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$$\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$$\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$$\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$$\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$

Q. A light source, which emits two wavelengths $\lambda_{1}=400 \mathrm{nm}$ and $\lambda_{2}=600 \mathrm{nm},$ is used in a Young’s double slit experiment. If recorded fringe widths for $\lambda_{1}$ and $\lambda_{2}$ are $\beta_{1}$ and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ respectively, then :-(A) $\beta_{2}>\beta_{1}$(B) $\mathrm{m}_{1}>\mathrm{m}_{2}$(C) From the central maximum, $3^{\mathrm{rd}}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$(D) The angular separation of fringes of $\lambda_{1}$ is greater than $\lambda_{2}$

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**Sol.**(A,B,C)$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$$\mathrm{B}_{2}>\beta_{1}$$\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$$\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$

Q. A Young’s double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $=4 / 3$ ) as shown in the figure. The positions of maxima on the surface of water are given by $x^{2}=p^{2} m^{2} \lambda^{2}-d^{2},$ where $\lambda$ is the wavelength of light in air (refractive index $=1$, $2 d$ is the separation between the slits and $m$ is an integer. The value of p is.

**[JEE Advanced 2015]**
Q. 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 $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ 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 = 3m from the mid-point of $\mathrm{S}_{1} \mathrm{S}_{2}$, 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 $\mathrm{S}_{1} \mathrm{S}_{2}$. Which of the following is (are) true of the intensity pattern on the screen ?(A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction(B) Semi circular bright and dark bands centered at point O(C) The region very close to the point O will be dark(D) Straight bright and dark bands parallel to the x-axis

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**Sol.**(B,C)Path difference at point O = d = .6003 mm = 600300 nm$=\frac{2001}{2}(600 \mathrm{nm})=1000 \lambda+\frac{\lambda}{2}$$\Rightarrow$ minima form at point $\mathrm{O}$Line $S_{1} S_{2}$ and screen are $\perp$ to each other so fringe pattern is circular (semi-circular because only half of screen is available)

Q. Two coherent monochromatic point sources $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of wavelength $\lambda$ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct ?(A) A dark spot will be formed at the point $\mathrm{P}_{2}$(B) The angular separation between two consecutive bright spots decreases as we move from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ along the first quadrant(C) At $\mathrm{P}_{2}$ the order of the fringe will be maximum(D) The total number of fringes produced between $P_{1}$ and $\mathrm{P}_{2}$ in the first quadrant is close to 3000

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**Sol.**(C,D)

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Excellent

Not so well explained

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Good questions.

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