In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t is ____ cm.

Solution & Explanation

### Related Formula textShift (x) = fracDd (mu - 1)t Alternatively written: d left(fracxDright) = (mu - 1)t ### Core Logic Given parameters: Distance between slits, d = 0.1text cm Distance to screen, D = 50text cm Fringe shift, x = 0.2text cm Refractive index, mu = 1.5 ### Step 1: Calculate Thickness Rearranging the formula for thickness t: t = fracx cdot dD(mu - 1) t = frac(0.2)(0.1)50(1.5 - 1) t = frac0.0250(0.5) = frac0.0225 t = 8 times 10^-4text cm ### Pattern Recognition Direct plug-and-play into the slab shift formula. Working completely in cm avoids unit conversion errors as long as all given lengths (x, d, D) are consistently in cm and the answer is requested in cm. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

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Wave Optics

More Wave Optics Previous-Year Questions — Page 2

Q2 jee_main_2025_07_april_morning Superposition and Interference
Two plane polarized light waves combine at a certain point whose electric field components are mathrmE_1 = mathrmE_0 sin omega t mathrm E _ 2 = mathrm E _ 0 sin left(omega t + frac pi3right) Find the amplitude of the resultant wave.
  • A. 0.9mathrmE
  • B. mathrmE_0
  • C. 1.7mathrmE_0
  • D. 3.4mathrmE_0

Solution

### Related Formula For two waves of identical direction and frequency superimposing with phase difference phi: E_textres = sqrtE_1^2 + E_2^2 + 2E_1E_2cosphi ### Core Logic The amplitudes of the two waves are E_1 = E_0 and E_2 = E_0. The phase difference is: phi = fracpi3 Substitute these values into the resultant amplitude equation: E_textres = sqrtE_0^2 + E_0^2 + 2E_0^2cosleft(fracpi3right) ### Step 1: Simplify the calculation Since cosleft(fracpi3right) = 0.5: E_textres = sqrt2E_0^2 + 2E_0^2(0.5) = sqrt3E_0^2 = sqrt3E_0 approx 1.732E_0 This is closest to 1.7mathrmE_0. ### Pattern Recognition Sees: Equal amplitudes (A) with a phase angle of 60^circ (pi/3). Shortcut: The vector sum of two vectors of equal magnitude A separated by 60^circ is always sqrt3A approx 1.73A. If separated by 120^circ, it is A. If 90^circ, it is sqrt2A. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics
Q18 jee_main_2025_08_april_evening Young's Double Slit Experiment
In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case:
  • A. There shall be an interference pattern for red distinct from that for green.
  • B. There shall be no interference fringes.
  • C. There shall be alternate interference fringes of red and green.
  • D. There shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.

Solution

### Related Formula For a stable, visible interference pattern to form, the light sources passing through the two slits must be **coherent**: - They must have the same wavelength (or frequency). - They must maintain a constant phase difference over time. ### Core Logic If one slit is covered by a red filter and the other by a green filter: - Only red light (lambda_textred approx 700mathrm~nm) passes through the first slit. - Only green light (lambda_textgreen approx 500mathrm~nm) passes through the second slit. Since the two passing waves have completely different wavelengths and frequencies, they are **incoherent**. ### Step 1: Resulting Pattern Analysis Coherent sources are a prerequisite for producing stable bright and dark interference fringes. Incoherent waves of different frequencies merely superimpose to create a general background illumination without any distinct, observable spatial fringe lines. Thus, **there shall be no interference fringes**. ### Pattern Recognition Sees: YDSE + opposite colored filters (red and green) on slits. Trap: Don't get confused thinking separate patterns will overlay. Since the slits emit different colors, the sources are incoherent, so the interference term langle cosphi rangle = 0. Shortcut: Different colors = different wavelengths = incoherent sources = NO fringes. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics
Q25 jee_main_2025_28_jan_morning Young's Double Slit Experiment
A double slit interference experiment performed with a light of wavelength 600mathrmnm forms an interference fringe pattern on a screen with 10^mathrmth bright fringe having its centre at a distance of 10mathrmmm from the central maximum. Distance of the centre of the same 10^mathrmth bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660mathrmnm would be ____________________
Numerical Answer. Answer: 11 to 11

Solution

### Related Formula mathrmY = fracmathrmnlambda mathrmDmathrmd implies mathrmY propto lambda ### Core Logic Since the fringe index mathrmn and apparatus parameters mathrmD, mathrmd remain constant across both runs: fracmathrmy_2mathrmy_1 = fraclambda_2lambda_1 Substituting the values into the proportionality equation: fracmathrmy_210 mathrm~mm = frac660 mathrm~nm600 mathrm~nm mathrmy_2 = 10 times 1.1 = 11 mathrm~mm ### Step 1: Final Numerical Value The distance of the tenth bright fringe shifts to exactly 11 mathrm~mm. ### Pattern Recognition Fringe position scales linearly with wavelength in standard Young's setups. Increasing the wavelength by 10\% shifts the entire pattern outward by exactly 10\%. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics
Q23 jee_main_2025_03_april_morning Interference Intensities Difference
Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of x is ________.
Numerical Answer. Answer: 24 to 24

Solution

### Related Formula For superposition of two coherent beams of intensities I_1 and I_2: I_textmax = left(sqrtI_1 + sqrtI_2right)^2 I_textmin = left(sqrtI_1 - sqrtI_2right)^2 ### Core Logic Given values: - I_1 = 4I - I_2 = 9I Let's calculate the square roots of the intensities: - sqrtI_1 = sqrt4I = 2sqrtI - sqrtI_2 = sqrt9I = 3sqrtI ### Step 1: Calculating Max and Min Intensities Substitute these values into the intensity formulas: I_textmax = (2sqrtI + 3sqrtI)^2 = (5sqrtI)^2 = 25I I_textmin = (3sqrtI - 2sqrtI)^2 = (1sqrtI)^2 = I ### Step 2: Finding the Difference The difference between the maximum and minimum intensities is: I_textmax - I_textmin = 25I - I = 24I Since this difference is given as xI: x = 24 ### Pattern Recognition Algebraic Shortcut: I_textmax - I_textmin = left(sqrtI_1 + sqrtI_2right)^2 - left(sqrtI_1 - sqrtI_2right)^2 = 4sqrtI_1 I_2 Substitute I_1 = 4I and I_2 = 9I: 4sqrt4I cdot 9I = 4sqrt36 I^2 = 4 times 6I = 24I. This beautiful identity (4ab formula) lets you solve the problem instantly without separately calculating maximum and minimum values! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics
Q15 jee_main_2025_04_april_evening Polarisation
Two polarisers P_1 and P_2 are placed in such a way that the intensity of the transmitted light will be zero. A third polariser P_3 is inserted in between P_1 and P_2 at the particular angle between P_2 and P_3. The transmitted intensity of the light passing through all the three polarisers is maximum. The angle between the polarisers P_2 and P_3 is:
  • A. fracpi4
  • B. fracpi6
  • C. fracpi8
  • D. fracpi3

Solution

### Related Formula Malus's Law: I = I_0 cos^2theta ### Core Logic Since P_1 and P_2 are crossed, the angle between their transmission axes is 90^circ. Let the angle between P_1 and P_3 be theta. Then the angle between P_3 and P_2 is left(90^circ - thetaright). Intensity after passing through P_3: I_1 = I_0 cos^2theta. Intensity after passing through P_2: I_textnet = I_1 cos^2(90^circ - theta) = I_0 cos^2theta sin^2theta. ### Step 1: Maximize Net Intensity Rewrite the expression: I_textnet = fracI_04 [2sinthetacostheta]^2 = fracI_04 [sin(2theta)]^2 For maximum transmitted intensity, sin(2theta) = 1 implies 2theta = 90^circ implies theta = 45^circ = fracpi4. The angle between P_2 and P_3 is 90^circ - 45^circ = 45^circ = fracpi4.
Polarizer alignment schema
Polarizer alignment schema
Polarizer alignment schema
Polarizer alignment schema
Polarizer alignment schema
Polarizer alignment schema
### Pattern Recognition Inserting a polarization filter at exactly 45^circ ( pi/4) between crossed polarizers symmetrically splits up components, maximizing overall transmission throughput. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

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