Chapter 16: Problem 2219
A Plane mirror produces a magnification of (A) 0 (B) \(+1\) (C) \(-1\) (D) \(\infty\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 16: Problem 2219
A Plane mirror produces a magnification of (A) 0 (B) \(+1\) (C) \(-1\) (D) \(\infty\)
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeThe distance between the first and sixth minima in the diffraction pattern of a single slit, it is \(0.5 \mathrm{~mm}\). The screen is \(0.5 \mathrm{~m}\) away from the Slit. If the wavelength of light is \(5000 \AA\), then the width of the slit will be \(\mathrm{mm}\) (D) \(1.0\) (A) 5 (B) \(2.5\) (C) \(1.25\)
A concave lens forms the image of an object such that the distance between the object and the image is \(10 \mathrm{~cm}\) and the magnification produced is \((1 / 4)\), the focal length of lens will be \(\mathrm{cm}\) (A) - 6.2 (B) \(-12.4\) (C) \(-4.4\) (D) \(-8.8\)
It is difficult to see through the fog because (A) light is scattered by the droplets in the fog. (B) fog absorbs light. (C) refractive index of fog is infinity. (D) light suffers total internal refection.
The power of plane glass is (A) \(\infty\) (B) 0 (C) \(\overline{2 \mathrm{D}}\) (D) \(4 \mathrm{D}\)
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
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