The minimum number of vectors of unequal magnitude required to

Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following

Law Incorporation [45]

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

d sin θ= m λ constructive interference

d sin θ = (m + ½) λ destructive interference

in general this phenomenon occurs for small angles, for which we can write

tanθ = y / L

tan te = sin tea / cos tea = sin tea

sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

d y / L = m λ

the distance between maximum clos is and

y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

d y / L = (m + ½) κ

y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

7 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Discuss what would happen to electrostatic force acting between two charged particles if you did the following. A. Increased the

zimovet [89]

Answer:B. Increased the amount of charge.

Explanation:

3 0

2 years ago

What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still

Inessa [10]

Answer:

physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

3 0

3 years ago

Elliptical galaxies are frequently found a) Inside the Milky Way b) In galaxy clusters c) In the Galactic bulge d) In the Local

sergejj [24]

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

5 0

3 years ago