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Chapter 17: Problem 2402

An image of sun is formed by a lens of focal length \(30 \mathrm{~cm}\) on the metal surface of a photo-electric cell and a photoelectric current (I) is produced. The lens forming the image is then replaced by another of the same diameter but of focal length of \(15 \mathrm{~cm}\). The photoelectric current in this case is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) 1 (C) \(2 \mathrm{I}\) (D) \(4 \mathrm{I}\)

Short Answer

Expert verified
The photoelectric current in the second case is \(\frac{1}{4} I\), so the correct answer is (D) \(4 \mathrm{I}\).

Step by step solution

01

Determine the image distance for both lenses 1 and 2

To get the image distance for each lens, we need to remember that the Sun is considered to be at an infinite distance and for such an object, the image forms at the focal point. So, for lens 1 of focal length \(30\text{cm}\), the image distance is \(30\text{cm}\). For lens 2 of focal length \(15\text{cm}\), the image distance is \(15\text{cm}\).
02

Calculate the light intensity ratio

To calculate the light intensity ratio, we have to apply the inverse square law of light intensity for both lenses. Let the light intensities for lenses 1 and 2 be \(I_1 \text{ and } I_2\) respectively. Then, we have the relation: \[\frac{I_1}{I_2} = \frac{d_2^2}{d_1^2}\] where, \(d_1\) is the image distance for lens 1 (30 cm) and \(d_2\) is the image distance for lens 2 (15 cm). By substituting the values for the image distances, we get: \[\frac{I_1}{I_2} = \frac{15^2}{30^2}\]
03

Find the light intensity ratio

To find the light intensity ratio, we need to simplify the previous expression: \[\frac{I_1}{I_2} = \frac{(15)^2}{(30)^2} = \frac{1}{2^2}\] thus, \[\frac{I_1}{I_2} = \frac{1}{4}\]
04

Determine the photoelectric current ratio

We know that the photoelectric current is directly proportional to the light intensity. So, the intensity ratio calculated previously is the same as the photoelectric current ratio. Therefore, the ratio between the photoelectric current for lens 1 denoted as \(I\) and the photoelectric current for lens 2 denoted as \(I'\) is: \[\frac{I'}{I} = \frac{1}{4}\] Rewriting this expression in terms of \(I'\), we get: \[I' = \frac{1}{4} I\] From the given options, we can choose the correct answer as: (D) \(4 \mathrm{I}\)

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Most popular questions from this chapter

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)

In which of the following phenomena the photon picture is required? (A) Energy distribution in black body radiation (B) Compton scattering (C) Photoelectric effect (D) all of the above

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)
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