• Light that has a 650-nm wavelength is incident upon two narrow slits that are separated by 0.500 mm. An interference pattern from the slits is projected onto a screen that is 3.00 m away. (a) What is the separation distance on the screen of the first bright fringe from the central bright fringe? (b) What is the separation distance on the screen of the second dark fringe from the central bright fringe?

(a) 3.9 mm

(b) 5.85 mm

Explanation:

Given that:

d = 0.500 mm, we need to convert this to meters so, d = 5.0 × m

lambda = 650 nm, this has to be converted to meters too so,

lambda = 6.5 × m and L = 3.00 m

(the symbol for lambda is not available so I'm just using the worded form)

(a) the general formula for finding the interference pattern is

dsin(theta) = nlambda

where sin(theta) = since the angle is not provided and n = 1 here since it refers to first bright fringe

(5.0 × )() = (1)(6.5 × )

= [(3)(6.5 × )] ÷ (5.0 × )

= 0.0039 m = 3.9 mm

(b) for the dark fringe, the formula will now be

dsin(theta) = (n-0.5)lambda

(5.0 × )() = (2-0.5)(6.5 × )

= [(3)(1.5)(6.5 × )] ÷ (5.0 × )

= 0.00585 m = 5.85 mm

Hope that answers the question. Have a great day!

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000