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Chapter 18: Problem 151

Magnification of a compound microscope is \(30 .\) Focal length of eyepiece is \(5 \mathrm{~cm}\), and the image is formed at a distance of distinct vision of \(25 \mathrm{~cm}\). The magnification of objective lens is

Short Answer

Expert verified
The magnification of the objective lens is \(5\).

Step by step solution

01

Calculate the magnification of the eyepiece

The magnification of the eyepiece can be calculated using the formula: \(m_e = 1 + \frac{D}{f_e}\) Where: - \(m_e\) is the magnification of the eyepiece - \(D\) is the distance of distinct vision (25 cm) - \(f_e\) is the focal length of the eyepiece (5 cm) Substituting the given values, we get: \(m_e = 1 + \frac{25}{5}\)
02

Calculate the magnification of the objective lens

Now that we have the magnification of the eyepiece, we can use the magnification of the entire compound microscope to calculate the magnification of the objective lens using the following formula: \(M = m_o \times m_e\) Where: - \(M\) is the total magnification of the microscope (30) - \(m_o\) is the magnification of the objective lens (which we need to find) - \(m_e\) is the magnification of the eyepiece (calculated in step 1) Rearranging the formula to find the magnification of the objective lens, we get: \(m_o = \frac{M}{m_e}\) Substitute the values: \(m_o = \frac{30}{1 + \frac{25}{5}}\)
03

Calculate the magnification of the objective lens

Now, just perform the calculations: \(m_o = \frac{30}{1 + 5}\) \(m_o = \frac{30}{6}\) \(m_o = 5\) Thus, the magnification of the objective lens is \(5\).

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Most popular questions from this chapter

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 \(4200 \AA\)

In a converging lens of focal length \(f\) and the distance between real object and its real image is \(4 f\). If the object moves \(x_{1}\) distance towards lens, its image moves \(x_{2}\) distance away from the lens and when object moves \(y_{1}\) distance away from the lens its image moves \(y_{2}\) distance towards the lens, then choose the correct option (A) \(x_{1}>x_{2}\) and \(y_{1}>y_{2}\) (B) \(x_{1}y_{2}\) (D) \(x_{1}>x_{2}\) and \(y_{2}>y_{1}\)

A microscope has an objective of focal length \(1.5 \mathrm{~cm}\) and an eye- piece of focal length \(2.5 \mathrm{~cm}\). If the distance between objective and eye-piece is \(25 \mathrm{~cm}\), what is the approximate value of magnification produced for relaxed eye? (A) 75 (B) 110 (C) 140 (D) 25

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

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) \(20 \mathrm{~cm}\)
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