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Chapter 18: Problem 67

In a Young's double slit experiment, the slit separation is \(1 \mathrm{~mm}\) and the screen is \(1 \mathrm{~m}\) from the slit. For a monochromatic light of wavelength \(500 \mathrm{~nm}\), the distance of third minimum from the central maximum is (A) \(0.50 \mathrm{~mm}\) (B) \(1.25 \mathrm{~mm}\) (C) \(1.50 \mathrm{~mm}\) (D) \(1.75 \mathrm{~mm}\)

Short Answer

Expert verified
The distance of the third minimum from the central maximum in a Young's double slit experiment with given parameters is \(1.50 \mathrm{~mm}\). The correct answer is (C).

Step by step solution

01

Calculate the position of a minimum intensity on the screen

To find the position of a minimum on the screen, use the formula: $$x_m = \frac{m \cdot \lambda \cdot L}{d}$$ Where: - \(x_m\) is the position of the m-th minimum on the screen - \(m\) is the order of the minimum (in this case, \(m = 3\)) - \(\lambda\) is the wavelength of the light - \(L\) is the distance between the slits and the screen - \(d\) is the slit separation
02

Convert the given values to standard SI units

Slit separation (d) = \(1 \mathrm{~mm} = 1 \times 10^{-3} \mathrm{~m}\) Wavelength of monochromatic light (位) = \(500 \mathrm{~nm} = 500 \times 10^{-9} \mathrm{~m}\)
03

Calculate the position of the third minimum on the screen

Plug in the values into the formula to find the position of the third minimum on the screen. $$x_3 = \frac{3 \cdot (500 \times 10^{-9} \mathrm{~m}) \cdot 1 \mathrm{~m}}{1 \times 10^{-3} \mathrm{~m}}$$
04

Evaluate the result

Now, calculate the position of the third minimum. $$x_3 = \frac{3 \cdot (500 \times 10^{-9} \mathrm{~m})}{1 \times 10^{-3} \mathrm{~m}} = 1.5 \times 10^{-3} \mathrm{~m}$$
05

Express the result in the requested unit

Finally, convert the result back to the given unit (millimeters) and write it as an option. $$x_3 = 1.5 \times 10^{-3} \mathrm{~m} = 1.50 \mathrm{~mm}$$ So, the distance of the third minimum from the central maximum is 1.50 mm. The correct answer is (C).

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Most popular questions from this chapter

In an usual Young's double slit experiment with \(\lambda=5000 \AA, d=2 \mathrm{~mm}\), and \(D=1 \mathrm{~m}\), the number of maximum obtained on the screen if it is large enough to receive any number of fringes (A) 8001 (B) 6001 (C) 4001 (D) 2001

When an object is placed at a distance of \(25 \mathrm{~cm}\) from a mirror, the magnification is \(m_{1} .\) The object is moved \(15 \mathrm{~cm}\) away with respect to the earlier position along principal axis, magnification becomes \(m_{2}\). If \(m_{1} \times m_{2}=4\), the focal length of the mirror is (A) \(10 \mathrm{~cm}\) (B) \(30 \mathrm{~cm}\) (C) \(15 \mathrm{~cm}\) (D) \(20 \mathrm{~cm}\)

A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is (A) Circle (B) Hyperbola (C) Parabola (D) Straight line

A plano-convex lens has a thickness of \(4 \mathrm{~cm}\). When placed on a horizontal table with curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be \(3 \mathrm{~cm}\). If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of plane face is found to be \(25 / 8 \mathrm{~cm}\). The focal length of the lens is (A) \(50 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(100 \mathrm{~cm}\) (D) \(150 \mathrm{~cm}\)

Diameter of a plano-convex lens is \(6 \mathrm{~cm}\) and thickness at the centre is \(3 \mathrm{~mm}\). If speed of light in material of lens is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\), the focal length of the lens is \([2013]\) (A) \(20 \mathrm{~cm}\) (B) \(30 \mathrm{~cm}\) (C) \(10 \mathrm{~cm}\) (D) \(15 \mathrm{~cm}\)
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