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In a double slit experiment shown in figure, when light of wavelength 400 mathrm~nm is used, dark fringe is observed at P. If D = 0.2 mathrm~m, the minimum distance between the slits S_1 and S_2 is ________ mm.
Double slit experimental configuration layout mapping path differences to point P for Q56
The image outlines a symmetric source alignment where path lengths from Source to upper slit S1 and lower slit S2 are drawn, passing over distance scale components D and converging at a focus node P.

Numerical Answer Type:
Enter a numerical value Answer: 0.2 to 0.2 +4 marks

Solution & Explanation

### Related Formula For a first order dark fringe (minima) to form, the net optical path difference between the interfering paths must be an odd multiple of half-wavelength: Delta x = fraclambda2 ### Core Logic From the symmetric setup diagram geometry: * Total path length 1 = textSource rightarrow S_1 rightarrow P = sqrtD^2 + d^2 + sqrtD^2 + d^2 = 2sqrtD^2 + d^2 * Total path length 2 = textSource rightarrow S_2 rightarrow P = D + D = 2D Hence, the exact geometric path difference is: Delta x = 2sqrtD^2 + d^2 - 2D ### Step 1: Set up Path Difference Balance Equating path difference to the minima criterion: 2sqrtD^2 + d^2 - 2D = fraclambda2 sqrtD^2 + d^2 - D = fraclambda4 implies sqrtD^2 + d^2 = D + fraclambda4 ### Step 2: Expand and Isolate Slit Space d Squaring both sides: D^2 + d^2 = D^2 + fraclambda^216 + frac2 D lambda4 d^2 = fracD lambda2 + fraclambda^216 ### Step 3: Approximate and Substitute Values Since lambda (400 mathrm~nm) is tiny compared to D (0.2 mathrm~m), the term fraclambda^216 is completely negligible: d^2 approx fracD lambda2 Substituting values: d^2 = frac0.2 times 400 times 10^-92 = 400 times 10^-10 mathrm~m^2 d = sqrt400 times 10^-10 = 20 times 10^-5 mathrm~m = 0.20 times 10^-3 mathrm~m = 0.20 mathrm~mm Therefore, the minimum distance between the slits is 0.20 mathrm~mm. ### Pattern Recognition This non-standard double slit setup calculates the primary total path directly from geometric lines rather than relying on standard angular small angle assumptions (y d / D). Recognizing when secondary higher-order differentials (lambda^2) can be safely dropped keeps calculations clean. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

Reference Study Guides

More Wave Optics Previous-Year Questions — Page 5

Q40 jee_main_2024_30_jan_morning Diffraction from a Single Slit
The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mathrm~mm is focused on the focal plane of a convex lens of focal length 100 mathrm~cm. The width of the 1^mathrmst secondary maxima will be :
  • A. 2 mathrm~mm
  • B. 2 mathrm~cm
  • C. 0.02 mathrm~mm
  • D. 0.2 mathrm~mm

Solution

### Related Formula textWidth of secondary maxima = fraclambda Da ### Core Logic In a single slit diffraction pattern, the linear width of any secondary maxima (fringe width of secondary bright bands) is given by W = fraclambda Da, whereas the central maximum is double this width (2fraclambda Da). ### Step 1: Parameter Identification Given values: Slit width, a = 0.2 times 10^-3 mathrm~m Wavelength, lambda = 400 times 10^-9 mathrm~m Distance to screen (focal length of the lens), D = 100 times 10^-2 mathrm~m = 1 mathrm~m ### Step 2: Execution Substitute these into the formula: textWidth = frac400 times 10^-90.2 times 10^-3 times 1 textWidth = frac4000.2 times 10^-6 mathrm~m textWidth = 2000 times 10^-6 mathrm~m = 2 times 10^-3 mathrm~m textWidth = 2 mathrm~mm ### Pattern Recognition Remember to strictly distinguish between central maximum (2lambda D/a) and secondary maxima (lambda D/a). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics
Q36 jee_main_2024_31_jan_evening Polarization by Reflection (Brewster's Law)
When unpolarized light is incident at an angle of 60^circ on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is
  • A. 30^circ
  • B. 60^circ
  • C. 90^circ
  • D. 45^circ

Solution

### Related Formula Brewster's Law states that at complete polarization upon reflection, the reflected and refracted rays are perpendicular to each other: i_p + r = 90^circ ### Core Logic The incident angle is given as i_p = 60^circ. At this angle, since the reflected ray is completely polarized, the geometry of Brewster's angle applies.
Polarization by Reflection (Brewster's Law) diagram for Q36 - JEE Main 2024 Evening
Polarization by Reflection (Brewster's Law) diagram for Q36 - JEE Main 2024 Evening
### Step 1: Calculate Refraction Angle 60^circ + r = 90^circ r = 90^circ - 60^circ = 30^circ ### Pattern Recognition The condition "reflected ray is completely polarized" is a direct trigger for Brewster's Law (i_p + r = 90^circ). No refractive index (mu) calculation is needed if only the geometric angle is asked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics Class 12 Physics: Ray Optics and Optical Instruments
Q56 jee_main_2024_31_jan_morning Interference Of Waves
Two waves of intensity ratio 1:9 cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is I_1 (b) Waves are coherent is I_2 and differ in phase by 60^circ. If fracI_1I_2 = frac10x then x =
Numerical Answer. Answer: 13 to 13

Solution

### Related Formula I_textincoherent = I_A + I_B I_textcoherent = I_A + I_B + 2sqrtI_A I_B cosphi ### Core Logic Let the individual intensities be I_A = I_0 and I_B = 9I_0. For incoherent waves, the net intensity is simply the algebraic sum: I_1 = I_A + I_B = I_0 + 9I_0 I_1 = 10I_0 ### Step 2: Coherent Waves Interference For coherent waves with a phase difference of phi = 60^circ: I_2 = I_A + I_B + 2sqrtI_A I_B cos(60^circ) I_2 = I_0 + 9I_0 + 2sqrt(I_0)(9I_0) left(frac12right) I_2 = 10I_0 + 2(3I_0) left(frac12right) I_2 = 10I_0 + 3I_0 = 13I_0 ### Step 3: Finding x Taking the ratio: fracI_1I_2 = frac10I_013I_0 = frac1013 Comparing with the given expression frac10x, we get: x = 13 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

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