Problem 108 - Master Resource Book in JEE Main Physics

Question

A car is fitted with a convex side-view mirror of focal length \(20 \mathrm{~cm}\). A second car \(2.8 \mathrm{~m}\) behind the first car is overtaking the first car at a relative speed of \(15 \mathrm{~m} / \mathrm{s}\). The speed of the image of he second car as seen in the mirror of the first one is (a) \(\frac{1}{15} \mathrm{~m} / \mathrm{s}\) (b) \(10 \mathrm{~m} / \mathrm{s}\) (c) \(15 \mathrm{~m} / \mathrm{s}\) (d) \(\frac{1}{10} \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

(d) \(\frac{1}{10} \text{ m/s}\).

Step 1: Understand the mirror formula for a convex mirror

In a convex mirror, the mirror formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Note that for a convex mirror, both \( f \) and \( v \) are positive while \( u \) is negative since the object distance is measured against the direction of incoming light.

Step 2: Assign known values

In this problem, the focal length \( f = 20 \text{ cm} = 0.2 \text{ m} \) (converted to meters for consistency with other units), and the object distance \( u = -2.8 \text{ m} \). We will use these values to find the image distance \( v \).

Step 3: Use the mirror formula to find image distance \( v \)

Using the mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Substituting the given values: \( \frac{1}{0.2} = \frac{1}{v} + \frac{1}{-2.8} \). Solving for \( \frac{1}{v} \), we get: \( \frac{1}{v} = \frac{1}{0.2} - \frac{1}{2.8} = 5 - \frac{1}{2.8} = \frac{14 - 1}{2.8} = \frac{13}{2.8} \). Thus, \( v = \frac{2.8}{13} \text{ m} \).

Step 4: Differentiate to find the speed of the image

We need the rate of change of image distance \( \frac{dv}{dt} \). Differentiate the mirror formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) with respect to time \( t \): \( 0 = \frac{-1}{v^2} \frac{dv}{dt} + \frac{-1}{u^2} \frac{du}{dt} \). Now substitute \( \frac{du}{dt} = 15 \text{ m/s} \) (since the car is moving towards the mirror): \( 0 = \frac{-1}{v^2} \frac{dv}{dt} - \frac{1}{2.8^2} \times 15 \). Solving gives: \( \frac{dv}{dt} = \frac{15 \cdot (\frac{v}{u})^2}{1} \). Substitute \( u = -2.8 \text{ m} \) and \( v = \frac{2.8}{13} \) to find \( \frac{dv}{dt} = \frac{15 \cdot \left( \frac{\frac{2.8}{13}}{-2.8} \right)^2}{1} = \frac{15 \cdot \frac{1}{169}}{1} = \frac{15}{169} \text{ m/s} \).

Step 5: Choose the closest answer

Since \( \frac{15}{169} \text{ m/s} \approx \frac{1}{10} \text{ m/s} \), which is closest to the given option (d), the answer is \( \frac{1}{10} \text{ m/s} \).

Key Concepts

Mirror Formula

In the world of optics, the mirror formula is a fundamental relationship that describes the geometry of image formation by spherical mirrors. For a convex mirror, this formula is expressed as \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Here, \( f \) represents the focal length, \( v \) the image distance, and \( u \) the object distance.

In the context of convex mirrors, a critical point to remember is that both the focal length \( f \) and image distance \( v \) are positive, as they form a virtual image on the same side as the incident light. However, the object distance \( u \) is negative because the object is positioned in the direction opposite to the incident light. This understanding of sign conventions is essential for correctly applying the mirror formula, particularly when calculating where an image will appear relative to the mirror.

Image Distance

The image distance \( v \) in optics refers to the distance from the mirror to the image that is formed. In a convex mirror, this image is virtual, meaning it cannot be projected onto a screen but is seen as though it is located behind the mirror.

To determine the image distance using the mirror formula, you need the focal length \( f \) and the object distance \( u \). As seen in the original exercise, we start with the equation \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). By substituting the known values for \( f \) and \( u \) into this formula, we can solve for \( v \).

• Example: With \( f = 0.2 \text{ m} \) and \( u = -2.8 \text{ m} \), substituting these gives \( \frac{1}{0.2} = \frac{1}{v} + \frac{1}{-2.8} \).
• Solving for \( \frac{1}{v} \) enables us to find \( v = \frac{2.8}{13} \text{ m} \).

Understanding how to calculate image distance is pivotal for predicting how the object's image will be viewed, whether it will be magnified, reduced, or the same size.

Relative Speed

The concept of relative speed is used to determine how quickly two objects are moving in relation to each other. In the context of the exercise, it refers to the speed at which the second car appears to overtake the first car.

When observing the second car from the first car's side-view mirror, relative speed helps us to calculate the speed at which the image of the second car moves within the convex mirror. The object distance \( u \) is changing due to the car's movement, and this change plays a crucial role in determining the image's speed.

• Example: If the second car moves towards the first car at \( 15 \text{ m/s} \), the relative speed helps in finding the rate of change of the image distance \( \frac{dv}{dt} \).
• This concept is key when calculating \( \frac{dv}{dt} \), as seen in the exercise, enabling us to find how the image position shifts over time.

Understanding relative speed is essential for seeing how dynamic situations affect what you observe in the mirror.

Differentiation in Optics

Differentiation in optics is a mathematical tool used to find how quickly something changes with respect to another variable—in this case, time. When an object is moving, its distance from the mirror changes, affecting the image distance. Differentiation helps in determining the speed at which this change occurs.

In the exercise, we differentiate the mirror formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) with respect to time \( t \) to find the rate at which the image distance \( v \) changes, \( \frac{dv}{dt} \). Consider the steps:

• First, acknowledge that \( \frac{du}{dt} \) is given by the car's speed, which is \( 15 \text{ m/s} \).
• Use differentiation to set up the equation: \( 0 = \frac{-1}{v^2} \frac{dv}{dt} + \frac{-1}{u^2} \frac{du}{dt} \).
• Substitute values and solve for \( \frac{dv}{dt} \), achieving a result in terms of \( \text{m/s} \).

This process of differentiation helps us understand the dynamic relationship between moving objects and their virtual images.