Is distance to screen (L) directly proportional to fringr space or inversly
Added by Jason S.
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Fringe spacing (β) is given by the formula: \[ \beta = \frac{\lambda L}{d} \] where: - \(\lambda\) = wavelength of light - \(L\) = distance to the screen - \(d\) = distance between the slits Show more…
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For double-slit interference, if the distance from slits to the screen, L, is decreased to L/2. Suppose L>>d, where d is the distance between two slits. The distance between adjacent fringes on the screen... a. increases by a factor of 4 b. decreases by a factor of 4 c. decreases by a factor of 2 d. increases by a factor of 2 e. depends on the width of the slits
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A lens is placed on a flat plate of glass to test whether its surface is spherical. (a) Show that the radius $r_{m}$ of the $m$ th dark ring should be $$ r_{m}=\sqrt{m \lambda R} $$ where $R$ is the radius of curvature of the lens surface facing the plate and the wavelength of the light used is $\lambda$. Assume that $r_{m} \ll R .$ [Hint: Start by finding the thickness $t$ of the air gap at a radius $r=R \sin \theta=R \theta$. Use small angle approximations.] (b) Are the dark fringes equally spaced? If not, do they get closer together or farther apart as you move out from the center?
A lens is placed on a flat plate of glass to test whether its surface is spherical. (a) Show that the radius $r_{m}$ of the mth dark ring should be $$ r_{m}=\sqrt{m \lambda R} $$ where $R$ is the radius of curvature of the lens surface facing the plate and the wavelength of the light used is $\lambda$ Assume that $r_{m}<<R .$ [Hint: Start by finding the thickness $t$ of the air gap at a radius $r=R \sin \theta=R \theta .$ Use small-angle approximations.] (b) Are the dark fringes equally spaced? If not, do they get closer together or farther apart as you move out from the center? (FIGURE CANNOT COPY)
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