0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
To know more about Fringes refer to: brainly.com/question/15649748
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As per as my knowledge
The speed of a wave in a medium is affected by <u>d</u><u>e</u><u>n</u><u>s</u><u>i</u><u>t</u><u>y</u>,<u> </u><u>w</u><u>a</u><u>v</u><u>e</u><u>l</u><u>e</u><u>n</u><u>g</u><u>t</u><u>h</u> and <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u>:)
(Good luck on your test and mark me brainliest if this helps)
Answer:

Explanation:
Given: The latent heat of fusion for Aluminum is 
mass to be malted m = 0.75 Kg
Energy require to melt E = mL

Therefore, energy required to melt 0.75 Kg aluminum

Answer:
The longest wavelength of light is 666.7 nm
Explanation:
The general form of the grating equation is
mλ = d(sinθi + sinθr)
where;
m is third-order maximum = 3
λ is the wavelength,
d is the slit spacing (m/slit)
θi is the incident angle
θr is the diffracted angle
Note: at longest wavelength, sinθi + sinθr = 1
λ = d/m
d = 1/500 slits/mm
λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm
Therefore, the longest wavelength of light is 666.7 nm