Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be_______I.
Correct Answer: 24
Solution and Explanation
\(I_A = (\sqrt{I_1} + \sqrt{I_2})^2 = 25I\)
\(I_B = (\sqrt{I_1} - \sqrt{I_2})^2 = I\)
So,\(I_A - I_B = 25I - I\)
\(= 24I\)
So, the answer is 24.
Top Questions on Photoelectric Effect
- The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope of \( \frac{h}{e} \). Explain this observation.
- CBSE CLASS XII - 2025
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- Photoelectric Effect
- Explain the following observations using Einstein’s photoelectric equation:
(a) Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
(b) It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
(c) The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope \( \frac{h}{e} \).
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- Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): For monochromatic incident radiation, the emitted photoelectrons from a given metal have speed ranging from zero to a certain maximum value.
Reason (R): Each metal has a definite work function.
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- Photoelectric Effect
- The threshold frequency for a given metal is \( 3.6 \times 10^{14} \) Hz. If monochromatic radiations of frequency \( 6.8 \times 10^{14} \) Hz are incident on this metal, find the cut-off potential for the photoelectrons.
Given: - Threshold frequency, \( \nu_0 = 3.6 \times 10^{14} \) Hz
- Frequency of incident radiation, \( \nu = 6.8 \times 10^{14} \) Hz
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \)
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- Physics
- Photoelectric Effect
Einstein's Explanation of the Photoelectric Effect:
Einstein explained the photoelectric effect on the basis of Planck’s quantum theory, where light travels in the form of small bundles of energy called photons.
The energy of each photon is hν, where:- ν is the frequency of the incident light
- h is Planck’s constant
The number of photons in a beam of light determines the intensity of the incident light.When a photon strikes a metal surface, it transfers its total energy hν to a free electron in the metal.A part of this energy is used to eject the electron from the metal, and this required energy is called the work function.The remaining energy is carried by the ejected electron as its kinetic energy.
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- Physics
- Photoelectric Effect
Questions Asked in JEE Main exam
- Let \( [x] \) denote the greatest integer function, and let \( m \) and \( n \) respectively be the numbers of the points, where the function \( f(x) = [x] + |x - 2| \), \( -2<x<3 \), is not continuous and not differentiable. Then \( m + n \) is equal to:
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- Differential Equations
- Let \( f(x) = \frac{x - 5}{x^2 - 3x + 2} \). If the range of \( f(x) \) is \( (-\infty, \alpha) \cup (\beta, \infty) \), then \( \alpha^2 + \beta^2 \) equals to.
- JEE Main - 2025
- Functions
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
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- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
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Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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Concepts Used:
Photoelectric Effect
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Photoelectric Effect Formula:
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,
- h is Planck’s constant.
- ν is the frequency of the incident photon.
- W is a work function.
- E is the maximum kinetic energy of ejected electrons: 1/2 mv².
Laws of Photoelectric Effect:
- The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
- There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
- The maximum kinetic energy of the photoelectrons increases with the increase in the frequency (provided frequency γ > γ Th exceeds the threshold limit) of the incident light. The maximum kinetic energy is free from the intensity of light.
- The process of photo-emission is an instantaneous process.