When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 8.80 cm wide on a screen that is 2.75 m away. How wide is the slit?

sinT= lambda/a

a= lambda/sinT

Isn't TanT=.0044/2.75 ? Since this is a very small angle, sinT appx= tanT

calculate slit width a.

To find the width of the slit, we can use the concept of diffraction. The equation that relates the width of the central diffraction peak to the width of the slit is given by:

w = (λ * D) / S

where:
w is the width of the central diffraction peak,
λ is the wavelength of the light (415 nm or 415 * 10^-9 m),
D is the distance between the slit and the screen (2.75 m), and
S is the width of the slit (what we want to find).

First, let's convert the wavelength from nanometers to meters:
λ = 415 * 10^-9 m

Now, we can substitute the values into the equation and solve for S:
w = (415 * 10^-9 m * 2.75 m) / S

We're given that the width of the central diffraction peak (w) is 8.80 cm or 8.80 * 10^-2 m. Substituting this value, we get:
8.80 * 10^-2 m = (415 * 10^-9 m * 2.75 m) / S

To find the width of the slit, S, we can rearrange the equation to solve for S:
S = (415 * 10^-9 m * 2.75 m) / (8.80 * 10^-2 m)

Calculating the value, we get:
S = 1.304 * 10^-3 m

Therefore, the width of the slit is approximately 1.304 mm.