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Chapter 16: Problem 2223

The power of plane glass is (A) \(\infty\) (B) 0 (C) \(\overline{2 \mathrm{D}}\) (D) \(4 \mathrm{D}\)

Short Answer

Expert verified
The power of plane glass is 0 (option B) since its parallel surfaces do not change the direction of incoming light rays, meaning it doesn't have any converging or diverging power.

Step by step solution

01

Understand Lens Power and Plane Glass Properties

Lens power is the property of a lens or optical system to converge or diverge incoming parallel rays of light. It is measured in Dioptres (D) and can be calculated using the formula: Power (P) = \(\frac{1}{f}\) where f is the focal length of the lens in meters. Plane glass has parallel surfaces which do not change the direction of incoming light rays, meaning it doesn't have any converging or diverging power.
02

Calculate the Power of Plane Glass

Since plane glass doesn't have any converging or diverging power, its focal length can be considered to be infinite. Plugging this into the power formula: Power (P) = \(\frac{1}{\infty}\)
03

Simplify the Result

Simplifying the expression: Power (P) = 0
04

Choose the Correct Option

Comparing the result with the given options, we can conclude that the correct answer is: (B) 0

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Most popular questions from this chapter

A ray of light is incident normally on one of the faces of a solid prism of apex angle \(30^{\circ}\) and refractive index \(\sqrt{2}\). The angle of minimum deviation is (A) \(39^{\circ}\) (B) \(42^{\circ}\)

Light of wave-length \(\lambda\) is incident on a slit of width \(\mathrm{d}\). The resulting diffraction pattern is observed on a screen placed at a distance \(\mathrm{D}\). The linear width of the principal maximum is equal to the width of the slit, then \(\mathrm{D}=\) (A) \(\left(\mathrm{d}^{2} / 2 \lambda\right)\) (B) \(\left(2 \lambda^{2} / \mathrm{d}\right)\) (C) \((\mathrm{d} / \lambda)\) (D) \((2 \lambda / \mathrm{d})\)

For four different transparent medium $\mathrm{n}_{41} \times \mathrm{n}_{12} \times \mathrm{n}_{21}=$ (A) \(\left(1 / \mathrm{n}_{41}\right)\) (B) \(\mathrm{n}_{41}\) (C) \(\mathrm{n}_{14}\) (D) \(\left(1 / \mathrm{n}_{14}\right)\)

Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given by $\mathrm{x}=\mathrm{n} \lambda(\mathrm{D} / \mathrm{d})$ and the position of dark fringes is given by \(\mathrm{x}=(2 \mathrm{n}-1)(N 2)(\mathrm{D} / \mathrm{d})\) where \(\mathrm{n}=1,2,3 \ldots \ldots \ldots \ldots\) for first second, third bright/dark fringe. The center of the fringe pattern is bright (for \(\mathrm{n}=0\) ). The width of each bright/dark fringe is $\beta=(\lambda \mathrm{D} / \mathrm{d})\(, Where \)\lambda=5000 \AA\(. Slits are \)0.2 \mathrm{~cm}$ apart and \(\mathrm{D}=1 \mathrm{~m}\) (i) If light of wavelength \(6000 \AA\) be used in the above experiment the fringe width would be \(\mathrm{mm}\) (A) \(0.30\) (B) 3 (C) \(0.6\) (D) 6 (ii) with the light of wavelength \(5000 \AA\), If experiment were carried out under water of a \(n=(4 / 3)\) the fringe width would be (A) zero (B) \((4 / 3)\) times (C) (3/4) times (D) none of these

The width of a single slit, if the first minimum is observed at an angle of \(2^{\circ}\) with a wavelength of light \(6980 \AA\) is \(\mathrm{mm}\) (A) \(0.2\) (B) \(2 \times 10^{-5}\) (C) \(2 \times 10^{5}\) (D) \(0.02\)
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