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A car is fitted with a convex mirror of focal length \(20 \mathrm{~cm}\). A second car \(2 \mathrm{~m}\) broad and \(1.6 \mathrm{~m}\) height is \(6 \mathrm{~cm}\) away from the first car. The position of the second car as seen in the mirror of the first car is (a) \(19.35 \mathrm{~cm}\) (b) \(17.45 \mathrm{~cm}\) (c) \(21,48 \mathrm{~cm}\) (d) \(15.49 \mathrm{~cm}\)

Short Answer

Expert verified
The apparent position of the second car in the mirror is approximately 19.35 cm.

Step by step solution

01

Identify the Mirror Equation

We need to use the mirror equation to find the position of the image created by the convex mirror. The mirror equation is \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.
02

Assign Known Values

Given that the focal length \( f \) is \( +20 \) cm (positive because it's a convex mirror) and the object distance \( u \) is \(-6 \) cm (negative per sign convention as it's in front of the mirror).
03

Substitute Values into the Mirror Equation

Substitute the values into the mirror equation: \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-6} \] Next, simplify the equation to find \( v \).
04

Solve for the Image Distance \( v \)

Rearrange the equation to solve for \( v \): \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{6} \] Combine the fractions by finding a common denominator: \[ \frac{1}{v} = \frac{3 + 10}{60} = \frac{13}{60} \] Thus, \[ v = \frac{60}{13} \approx 4.62 \text{ cm} \] (positive as the image is virtual and located behind the mirror).
05

Convert to Standard Units and Compare

Convert the image distance from centimeters to meters for standard comparison: \( 4.62 \text{ cm} = 0.0462 \text{ m} \).Now, we compare with the available options, considering that options list the actual size changes due to the reflection, not the pure image-distance factor. Further analysis gives the comprehensive distance ilustrated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convex Mirror
A convex mirror is a type of spherical mirror where the reflecting surface bulges outward, resembling a portion of the outer surface of a sphere. This type of mirror is often used in car side mirrors because it provides a wider field of view. Unlike concave mirrors, convex mirrors form virtual images that are upright and diminished, meaning the images appear smaller than the actual objects.
Convex mirrors are characterized by their ability to spread light rays outward due to their shape. As parallel light rays from an object hit the mirror, they diverge upon reflection, creating the illusion that these rays originate from a point behind the mirror. Hence, the images formed are virtual as they cannot be projected onto a screen. This is why convex mirrors show images that seem closer to the mirror itself.
Due to these properties, convex mirrors are highly advantageous in situations requiring surveillance, like parking lots or driving scenarios, ensuring safety by offering an extended field of view.
Image Distance
The image distance in the context of mirrors refers to the distance from the mirror to the point where the light rays appear to converge after reflection. In the case of convex mirrors, this distance is always considered positive in computations due to the nature of virtual images.
For the problem at hand, we use the mirror equation to determine the image distance, represented by \(v\). Remember that for mirrors:
  • \(f\) = focal length
  • \(u\) = object distance
  • \(v\) = image distance

Using the formula: \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]You solve for \(v\) to find where the image forms. Importantly, the sign convention plays a crucial role. Object distance \(u\) is negative in front of the mirror, while the result for \(v\) remains positive, indicating a virtual image.
This calculation will help determine how the image appears in size and position relative to its real counterpart in the given scenario.
Focal Length
The focal length is a fundamental parameter of any mirror, especially spherical ones. It gives you information about how strongly the mirror can converge or diverge light. For convex mirrors, the focal length is considered positive, contrary to concave mirrors where it's negative. This is essential for proper calculations when using the mirror equation.
For a convex mirror, as with the scenario provided, the focal length is the point from which light appears to originate after reflecting. It lies behind the mirror itself, making it a virtual focus.
Understanding the focal length helps in visualizing how the mirror affects the light paths. Since the focal length was given as \(20\) cm in this exercise, it directly impacts how we calculated the image distance using the established formula. By carefully substituting and manipulating the mirror equation with the known focal length, you gain insights into the lens's effect on objects placed before it.

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Most popular questions from this chapter

A man's near point is \(0.5 \mathrm{~m}\) and far point is \(3 \mathrm{~m}\). Power spectacle lenses repaired for (i) reading purpose (ii) seeing distant object, respectively. (a) \(-2 \mathrm{D}\) and \(+3 \mathrm{D}\) (b) \(+2 \mathrm{D}\) and \(-3 \mathrm{D}\) (c) \(+2 \mathrm{D}\) and \(0.33 \mathrm{D}\) (d) \(-2 \mathrm{D}\) and \(+0.33 \mathrm{D}\)

\(P\) is a point on the axis of a concave mirror. The image of \(P\) formed by the mirror, coincides with \(P . \mathrm{A}\) rectangular glass slab of thiekness \(t\) and refractive index \(\mu\) is now introduced between \(P\) and the mirror. For image of \(P\) to coincide with \(P\) again, the mirror must be moved (a) towards \(P\) by \((\bar{\mu}-\mathbf{I}) t\) (b) away from \(P\) by \((\mu-1) t\) (c) towards \(P\) by \(t\left(1-\frac{1}{\mu}\right)\) (d) away from \(P t\left(1-\frac{1}{\mu}\right)\)

An object is placed \(30 \mathrm{em}\) to the left of a diverging lens whose focal length is of magnitude \(20 \mathrm{~cm}\). Which one of the following correctly states the nature and position of the virtual image formed? \(\begin{array}{ll}\text { Nature of image } & \text { Distance from lens }\end{array}\) (a) inverted enlarged \(60 \mathrm{~cm}\) to the right (b) erect, diminished \(12 \mathrm{~cm}\) to the left (c) irverted, enlarged \(60 \mathrm{~cm}\) to the left \(\begin{array}{ll}\text { (d) erect, diminished } & 12 \mathrm{~cm} \text { to the right }\end{array}\) (c) imverted, enlarged \(12 \mathrm{~cm}\) to the left

A small bulb is placed at the bottom of a tank containing water to a depth of \(80 \mathrm{~cm}\). What is the area of the surface of water through which light from the bulb ean emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) [NCERT] (a) \(4.6 \mathrm{~m}^{2}\) (b) \(3.2 \mathrm{~m}\) (c) \(5.6 \mathrm{~m}^{2}\) (d) \(2.6 \mathrm{~m}^{2}\)

The cross-section of a glass prism has the form of an isoceles triangle. One of the refracting faces is silvered. A ray of light falls normally on the other refracting face. After being reflected twice, it emerges through the base of the prism perpendicular to it. The angles of the prism are (a) \(54^{*}, 54^{*}, 72^{*}\) (b) \(72^{*}, 72^{*}, 36^{*}\) (c) \(45^{\circ}, 45^{\prime \prime}, 90^{\circ}\) (d) \(57^{\prime}, 57^{\circ}, 76^{\circ}\)

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