HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

17: Light Waves

with Solutions - page 4
Qstn# iv-21-a Prvs-QstnNext-Qstn
  • #21-a
    What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?
    Figure
    Ans : The phase of a light wave reflecting from a surface differs by '`` \pi ``' from the light directly coming from the source.

    Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of `` \pi ``, which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero. (b) Here, separation between two slits is `` 2d``.
    Wavelength of the light is `` \lambda ``.
    Distance of the screen from the slit is `` D``.
    Consider that the bright fringe is formed at position y. Then,
    path difference, `` ∆x=\frac{y\times 2d}{D}=n\lambda ``.
    After reflection from the mirror, path difference between two waves is `` \frac{\lambda }{2}``.
    `` \Rightarrow \frac{y\times 2d}{D}=\frac{\lambda }{2}+n\lambda ``
    `` \,\mathrm{\,For\,}\,\mathrm{\,first\,}or\,\mathrm{\,der\,},\,\mathrm{\,put\,}n=0.``
    `` \Rightarrow y=\frac{\lambda D}{4d}``
    Page No 382:
  • #21-b
    At what distance from O does the first maximum occur?
    Figure
    Ans : Here, separation between two slits is `` 2d``.
    Wavelength of the light is `` \lambda ``.
    Distance of the screen from the slit is `` D``.
    Consider that the bright fringe is formed at position y. Then,
    path difference, `` ∆x=\frac{y\times 2d}{D}=n\lambda ``.
    After reflection from the mirror, path difference between two waves is `` \frac{\lambda }{2}``.
    `` \Rightarrow \frac{y\times 2d}{D}=\frac{\lambda }{2}+n\lambda ``
    `` \,\mathrm{\,For\,}\,\mathrm{\,first\,}or\,\mathrm{\,der\,},\,\mathrm{\,put\,}n=0.``
    `` \Rightarrow y=\frac{\lambda D}{4d}``
    Page No 382: