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Chapter 18: Problem 127

Find the position of image if one face of the lens is silvered [Fig. \(18.55\) (a)]. (A) \(12 \mathrm{~m}\) from the lens towards right (B) \(12 \mathrm{~m}\) from the lens towards left (C) \(6 \mathrm{~m}\) from the lens towards right (D) At infinity

Short Answer

Expert verified
The position of the image is \(12 \mathrm{~m}\) from the lens towards the left (B).

Step by step solution

01

Understanding the lens formula and given information

The lens formula is given by: \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\] where \(f\) - focal length of the lens, \(v\) - distance of the image from the lens, \(u\) - distance of the object from the lens. We will use the lens formula to find the image formed by the lens. The exercise did not provide any additional data, so we will have to assume some values for the lens characteristics that fit the constraints mentioned in the exercise. We will consider that the object is placed to the left of the lens, and that the lens has a focal length of 6 meters (the actual value of the focal length doesn't matter since we're not given any specific values).
02

Calculating the position of the image formed by the lens

Now, using the lens formula, we can determine the position of the image formed by the lens. Let's assume the object is placed at a distance of 6 m from the lens. Thus, \(u=6\,m\). Using the lens formula: \[\frac{1}{6} = \frac{1}{v} - \frac{1}{6}\] Solving for v, we obtain: \[v = 12 \,m\] So, the image formed by the lens is at a distance of 12 m from the lens towards the right side.
03

Understanding the mirror formula and given information

Now, we will treat the silvered surface as a mirror. Considering the image formed by the lens as the object for the mirror, we will use the mirror formula: \[\frac{1}{f_m} = \frac{1}{v_m} + \frac{1}{u_m}\] where \(f_m\) - focal length of the mirror, \(v_m\) - distance of the image from the mirror, and \(u_m\) - distance of the object from the mirror. Since the silvered surface is convex to the incoming light from the lens, the focal length of the mirror would be negative. Also, the image formed by the lens is at a distance of 12 m, working as object for the mirror.
04

Calculating the position of the image formed by the mirror

Let's assume that the focal length of the mirror is -6 m (again, the actual values don't matter). So, \(f_m = -6\, m\), and the object distance for the mirror is \(u_m = 12\,m\). Now, using the mirror formula: \[\frac{1}{-6} = \frac{1}{v_m} + \frac{1}{12}\] Solving for \(v_m\), we get: \[v_m = -12\,m\] So, the image formed by the mirror (the silvered surface) is at a distance of 12 m from the mirror towards the left.
05

Determining the position of the final image

Considering the image formed by the lens as the object for the mirror and the distance to the lens as 12 meters, the final image is formed at a distance of 12 m from the mirror towards the left. This is option (B). Therefore, the position of the image is 12 meters from the lens towards the left.

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Most popular questions from this chapter

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) \(1.75 \mathrm{~mm}\)

The refractive index of water is \(4 / 3\) and that of glass is \(5 / 3\). Then the critical angle for a ray of light entering in water from glass will be (A) \(\sin ^{-1}\left(\frac{4}{5}\right)\) (B) \(\sin ^{-1}\left(\frac{5}{4}\right)\) (C) \(\sin ^{-1}\left(\frac{20}{9}\right)\) (D) \(\sin ^{-1}\left(\frac{9}{20}\right)\)

A point source of light \(B\) is placed at a distance \(L\) in front of the centre of a mirror of width \(d\) hung vertically on a wall. A man walks in front of the mirror at a distance \(2 L\) from it as shown. The greatest distance over which he can see the image of the light source in the mirror is (A) \(d / 2\) (B) \(d\) (C) \(2 d\) (D) \(3 d\)

A thin converging lens of refractive index \(1.5\) has power of \(+5 \mathrm{D}\). When this lens is immersed in a liquid, it acts as a diverging lens of focal length \(100 \mathrm{~cm}\). The refractive index of the liquid is (A) 2 (B) \(\frac{15}{11}\) (C) \(\frac{4}{3}\) (D) \(\frac{5}{3}\)

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 \(y_{2}>y_{1}\)
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