A Young’s double slit apparatus has the slit separation of 0.15mm and the distan

Question

A Young’s double slit apparatus has the slit separation of
0.15mm and the distance between the slits and the viewing
screen is 150 cm.
A) If monochromatic light of wavelength ? = 550x10-9 m passes
through the slits, calculate the distance between the central
(bright) fringe and the adjacent bright fringe.
B) Now imagine that you are using X-rays, ?=1x10–9 m instead of
visible light in this experiment. What would the slit separation
need to be in order to have the distance between these first two
bright fringes be 1cm?

Explanation / Answer

  a.   Frindge width ( width of onefringe)      =   D/ d         =   wavelength,   D   =   Distancebetween screen and slits       d   =   distancebetween slits             =   550* 10-9 * 1.5 m/ 0.15 * 10-3             =   5.50*10-3   m   =   5.5 mm    b.   given   =   1cm   =   0.01 m       0.01   =   1* 10-9 * 1.5 / d          d   =   1.5* 10-10 /10-2   =   =   150nm         =   wavelength,   D   =   Distancebetween screen and slits       d   =   distancebetween slits             =   550* 10-9 * 1.5 m/ 0.15 * 10-3             =   5.50*10-3   m   =   5.5 mm    b.   given   =   1cm   =   0.01 m       0.01   =   1* 10-9 * 1.5 / d          d   =   1.5* 10-10 /10-2   =   =   150nm             =   5.50*10-3   m   =   5.5 mm    b.   given   =   1cm   =   0.01 m       0.01   =   1* 10-9 * 1.5 / d          d   =   1.5* 10-10 /10-2   =   =   150nm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.