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Intensity in Young's experiment at a point with path difference lambda/6 (or lambda/3, lambda/4), given the maximum intensity

The same question, asked 8 times: 3 May 2023 S2, 9 May 2023 S2, 15 May 2023 S2*, 9 May 2024 S2*, 19 Apr 2025 S2*, 19 Apr 2026 S1, 19 Apr 2026 S2*, 13 May 2026 S1. Every wording below, oldest first.

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  1. Question 1

    3 May 2023 S2

    47
    In Young's double slit experiment using monochromatic light of wavelength 'λ\lambda', the intensity of light at a point on the screen where path difference 'λ\lambda' is K units. The intensity of light at a point where the path difference is λ6\frac{\lambda}{6} is
    cos⁡π6=sin⁡π3=32\cos\frac{\pi}{6}=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}

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  2. Question 2

    9 May 2023 S2

    36
    In Young's double slit experiment using monochromatic light of wavelength 'λ\lambda', the maximum intensity of light at a point on the screen is 'K' units. The intensity of light at a point where the path difference is 'λ6\frac{\lambda}{6}', is cos⁡60∘=sin⁡30∘=0.5,sin⁡60∘=cos⁡30∘=32\cos 60^\circ = \sin 30^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}

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  3. Question 3

    9 May 2023 S2

    10
    In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is λ\lambda (λ\lambda = wavelength) is β\beta. The intensity at a point where the path difference is λ3\frac{\lambda}{3}, will be cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}

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  4. Question 4

    15 May 2023 S2

    44
    In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is λ\lambda is xx units, λ\lambda being the wavelength of light used. The intensity at a point where the path difference is λ/4\lambda/4 will be (cos⁡2π=1, cos⁡π2=0\cos 2\pi = 1,\, \cos \frac{\pi}{2} = 0)

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  5. Question 5

    9 May 2024 S2

    3
    In Young's double slit experiment, the wavelength of light used is λ\lambda. The intensity at a point is II where path difference is λ4\frac{\lambda}{4}. If I0I_0 denotes the maximum intensity, then the ratio II0\frac{I}{I_0} is (sin⁡π4=cos⁡π4=12)\left(\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\right)

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  6. Question 6

    19 Apr 2025 S2

    21
    In Young's double slit experiment, the intensity on screen at a point where path difference is λ4\frac{\lambda}{4} is K2\frac{K}{2}. The intensity at a point when path difference is ' λ\lambda ' will be

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  7. Question 7

    19 Apr 2026 S1

    141
    In Young's double slit experiment, the wavelength of light used is λ\lambda. The intensity on the screen at a point for path difference 'λ\lambda' is 'XX'. The intensity at the point for path difference (λ6)\left(\dfrac{\lambda}{6}\right) is (cos⁡180∘=−1,cos⁡30∘=32)\left(\cos 180^\circ = -1, \cos 30^\circ = \dfrac{\sqrt{3}}{2}\right)

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  8. Question 8

    19 Apr 2026 S2

    142
    In biprism experiment, the maximum intensity is I0I_0. If the path difference between the two interfering waves is λ3\dfrac{\lambda}{3}, then intensity at the point on the screen is [sin⁡30∘=cos⁡60∘=0.5,sin⁡60∘=cos⁡30∘=3/2][\sin 30^\circ = \cos 60^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \sqrt{3}/2]

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  9. Question 9

    13 May 2026 S1

    143
    The wavelength of light used in Young's double slit experiment, is λ\lambda. The intensity at a point on the screen is I, where the path difference is λ6\dfrac{\lambda}{6}. If I0I_0 is the maximum intensity then the ratio I to I0I_0 is (cos⁡30∘=32)\left(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\right)

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