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Chapter 18: Problem 130

If we put \(v=V+f\) and \(u=U+f\), the mirror formula \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) becomes (A) \((V+f)(U+f)=f^{2}\) (B) \(V U=f^{2}\) (C) \((V-f)(U-f)=f^{2}\) (D) \(V U=2 f^{2}\)

Short Answer

Expert verified
The correct answer is (B) \(VU = f^2\).

Step by step solution

01

Substitute the given values

Replace \(v\) with \(V+f\) and \(u\) with \(U+f\) in the mirror formula: \[\frac{1}{V+f}+\frac{1}{U+f}=\frac{1}{f}\]
02

Find a common denominator

To simplify the expression on the left-hand side of the equation, we need to find a common denominator for the two fractions. The common denominator here would be \((V+f)(U+f)\). So, the equation becomes: \[\frac{U+f}{(V+f)(U+f)} + \frac{V+f}{(V+f)(U+f)} = \frac{1}{f}\]
03

Combine the fractions and simplify

Now, we can combine the two fractions on the left-hand side of the equation into one and simplify: \[\frac{(U + f) + (V + f)}{(V + f)(U + f)} = \frac{1}{f}\] This simplifies to: \[\frac{U + V + 2f}{(V + f)(U + f)} = \frac{1}{f}\]
04

Cross-multiply and simplify further

To get rid of the fractions and isolate the terms, we will cross-multiply both sides of the equation: \[(U+V+2f)(f)=(V+f)(U+f)\] Expanding and simplifying we get: \[Uf + Vf + 2f^2 = VU + Uf + Vf + f^2\] Subtracting \(Uf\) and \(Vf\) from both sides: \[2f^2 = VU + f^2\] Subtracting \(f^2\) from both sides, we get the final simplified form of the equation: \[VU = f^2\]
05

Identify the correct option

Now that we have the final simplified form of the equation, we can compare it to the given options and find the correct answer. The simplified equation \(VU = f^2\) matches with option (B). Therefore, the correct answer is (B) \(VU = f^2\).

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Most popular questions from this chapter

In YDSE distance between the slits plane and screen is \(1 \mathrm{~m}\) and distance between two slits is \(5 \mathrm{~mm}\). If slabs of thickness \(2 \mathrm{~mm}\) and \(1.5 \mathrm{~mm}\) having refractive index \(1.5\) and \(1.4\) are placed in front of two slits, the shift of central maximum will be (A) \(2 \mathrm{~m}\) (B) \(8 \mathrm{~cm}\) (C) \(20 \mathrm{~cm}\) (D) \(80 \mathrm{~cm}\)

The maximum intensity in Young's double slit experiment is \(I_{0}\). Distance between the slits is \(d=5 \lambda\), where \(\lambda\) is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance \(D=10 d\) ? (A) \(\frac{I_{0}}{2}\) (B) \(\frac{3 I_{0}}{4}\) (C) \(I_{0}\) (D) \(\frac{I_{0}}{4}\)

In Young's double slit experiment, double slit of separation \(0.1 \mathrm{~cm}\) is illuminated by white light. A coloured interference pattern is formed on a screen \(100 \mathrm{~cm}\) away. If a pinhole is located on this screen at a distance of \(2 \mathrm{~mm}\) from the central fringe, the wavelength in the visible spectrum which will be absent in the light transmitted through the pin hole are (A) \(5714 \AA\) and \(4444 \AA\) (B) \(6000 \AA\) and \(5000 \AA\) (C) \(5500 \AA\) and \(4500 \AA\) (D) \(5200 \AA\) and \(4200 \AA\)

What will be the angular width of central maximum in Fraunhofer diffraction when light of wavelength \(6000 \AA\) is used and slit width is \(12 \times 10^{-5} \mathrm{~cm} ?\)

A circular beam of light of diameter \(d=2 \mathrm{~cm}\) falls on a plane surface of glass. The angle of incidence is \(60^{\circ}\) and refractive index of glass is \(\mu=3 / 2\). The diameter of the refracted beam is (A) \(4.0 \mathrm{~cm}\) (B) \(3.0 \mathrm{~cm}\) (C) \(3.26 \mathrm{~cm}\) (D) \(2.52 \mathrm{~cm}\)
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