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Chapter 15: Problem 2198

Infrared radiations are detected by (A) spectrometer (B) bolometer (C) photocells (D) geiger tubes

Short Answer

Expert verified
The correct answer is (B) bolometer, as it is specifically designed to detect and measure infrared radiations.

Step by step solution

01

Understand Infrared Radiations

Infrared radiations are a part of the electromagnetic spectrum, and they have wavelengths longer than visible light but shorter than microwaves. They are mainly used for heating, thermal sensing and imaging, and remote controls.
02

Option A: Spectrometer

A spectrometer is an instrument that is used to separate and analyze the components of a substance according to their wavelengths. They are used in spectrophotometry and spectroscopy to identify materials, substances or even elements. While this instrument can be used to detect some wavelengths, it is not specifically designed for infrared radiations.
03

Option B: Bolometer

A bolometer is a device used to measure the energy of incident electromagnetic radiation, particularly in the infrared region of the spectrum. They are sensitive sensors that can detect small changes in temperature, which makes them perfect for detecting infrared radiations.
04

Option C: Photocells

Photocells are electronic devices that are sensitive to light or other forms of radiant energy. When light falls on a photocell, it generates an electric current or a voltage. Photocells are typically used in applications like solar panels and light detection, such as for automatic doors and automated lighting systems. They are not designed to detect infrared radiations specifically.
05

Option D: Geiger Tubes

A Geiger tube (also known as Geiger-Müller counter) is an instrument used to detect ionizing radiations such as alpha particles, beta particles, and gamma rays. It is not used for detecting infrared radiations. Based on the information provided for each option, we can now determine the correct answer.
06

Conclusion

The correct answer is (B) bolometer, as it is specifically designed to detect and measure infrared radiations.

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Most popular questions from this chapter

In electromagnetic spectrum, the visible light lie between (A) radiowaves and microwaves (B) ultraviolet rays and infrared rays (C) ultraviolet rays and \(\mathrm{x}\) rays (D) infrared rays and microwaves

A plane electromagnetic wave of frequency \(25 \mathrm{MHz}\) travels in free space along the \(\mathrm{x}\) direction. At a particular point in space and time \(\mathrm{E}^{-}=6.3 \mathrm{j} \wedge \mathrm{Vm}^{-1}\) then \(\mathrm{B}^{-}\) at this point is (A) \(2.1 \times 10^{-8}\) i \(\mathrm{T}\) (B) \(2.1 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (C) \(1.89 \times 10^{9} \mathrm{k} \wedge \mathrm{T}\) (D) \(2.52 \times 10^{-7} \mathrm{k} \wedge \mathrm{T}\)

In an electromagnetic wave, if the amplitude of magnetic field is $3 \times 10^{-10} \mathrm{~T}$, the amplitude of the associated electric field will be (A) \(9 \times 10^{-2} \overline{\mathrm{Vm}^{-1}}\) (B) \(3 \times 10^{-10} \mathrm{Vm}^{-1}\) (C) \(3 \times 10^{-2} \mathrm{Vm}^{-1}\) (D) \(1 \times 10^{-18} \mathrm{Vm}^{-1}\)

The maximum electric field in a plane electromagnetic wave is $900 \mathrm{NC}^{-1}\(. The wave is going in the \)\mathrm{x}$ direction and the electric field is in the y direction. The maximum magnetic field in the wave is \(\mathrm{T}\) (A) \(3 \times 10^{-8}\) (B) \(3 \overline{\times 10^{-6}}\) (C) \(27 \times 10^{-6}\) (D) $27 \times 10^{10}$

A plane electromagnetic wave of frequency \(25 \mathrm{MHz}\) travels in free space along the \(\mathrm{x}\) direction. At a particular point in space and time \(\mathrm{E}^{-}=6.3 \mathrm{j} \wedge \mathrm{Vm}^{-1}\) then \(\mathrm{B}^{-}\) at this point is (A) \(2.1 \times 10^{-8}\) i \(\mathrm{T}\) (B) \(2.1 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (C) \(1.89 \times 10^{9} \mathrm{k} \wedge \mathrm{T}\) (D) \(2.52 \times 10^{-7} \mathrm{k} \wedge \mathrm{T}\)
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