Problem 19 - Master Resource Book in JEE Main Physics

Question

A small candle, \(2.5 \mathrm{~cm}\) in size is placed at \(27 \mathrm{~cm}\) in front of a concave mirror of radius of curvature \(36 \mathrm{~cm}\). At what distance from the mirror should a sereen be placed in order to obtain a sharp image? Deseribe the nature and size of the image. If the candle is moved closer to the mirror, how would the sereen have to be moved? (a) \(54 \mathrm{~cm}\) (b) \(27 \mathrm{~cm}\) (c) \(28 \mathrm{~cm}\) (d) \(475 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The screen should be placed at \(54 \mathrm{~cm}\) for a sharp image, which is real, inverted, and magnified 2 times. Moving the candle closer increases \( v \), so the screen must be moved further away.

Step 1: Find the mirror's focal length

The focal length \( f \) of a concave mirror is given by \( f = \frac{R}{2} \) where \( R \) is the radius of curvature. Here, \( R = 36 \mathrm{~cm} \), so the focal length is \( f = \frac{36}{2} = 18 \mathrm{~cm} \).

Step 2: Use the mirror equation

The mirror equation is \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \( v \) is the image distance and \( u \) is the object distance. We know \( u = -27 \mathrm{~cm} \) (object distance is taken as negative for concave mirrors). Substitute \( f = 18 \mathrm{~cm} \) and \( u = -27 \mathrm{~cm} \) to find \( v \):\[ \frac{1}{18} = \frac{1}{v} + \frac{1}{-27} \]

Step 3: Solve for the image distance \( v \)

Rearrange the equation and solve for \( \frac{1}{v} \):\[ \frac{1}{v} = \frac{1}{18} - \frac{1}{27} \]Find a common denominator and simplify:\[ \frac{1}{v} = \frac{3}{54} - \frac{2}{54} = \frac{1}{54} \]Thus, \( v = 54 \mathrm{~cm} \).

Step 4: Describe the nature and size of the image

When \( v \) is positive, it indicates that the image is real and formed on the same side as the object. The magnification \( m \) can be found using \( m = -\frac{v}{u} \):\[ m = -\frac{54}{-27} = 2 \]This means the image is upright and magnified 2 times the size of the object. Since the object is \( 2.5 \mathrm{~cm} \), the image will be \( 2.5 \times 2 = 5.0 \mathrm{~cm} \).

Step 5: Effect of moving the candle closer to the mirror

As the object (candle) moves closer to the mirror, the object distance \( u \) becomes less negative, leading to a change in image distance \( v \) according to the mirror equation. To maintain a sharp image, the screen would need to be moved further away from the mirror, thus increasing \( v \).

Key Concepts

Understanding the Mirror Equation

The mirror equation is fundamental when working with spherical mirrors, like concave ones. It helps us determine the position and nature of the image formed by the mirror. The equation is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]This equation relates:

• f: the focal length of the mirror, which is positive for concave mirrors.
• v: the image distance from the mirror. If this is positive, the image is real and located on the same side as the object.
• u: the object distance from the mirror, which is taken as negative for concave mirrors.

In application, knowing two of these quantities allows you to solve for the third. For instance, if you are given the focal length and the object distance, you can calculate the image distance. This is especially useful in understanding where to place a screen to capture a sharp image. Learning how to manipulate this equation through simple algebra can drastically aid your understanding of mirror behavior.

Calculating the Focal Length of a Concave Mirror

Calculating the focal length of a concave mirror is straightforward when you know the mirror's radius of curvature, denoted as \( R \). The formula to find the focal length, \( f \), is:
\[ f = \frac{R}{2} \]In the given problem, \( R = 36 \mathrm{~cm} \). Thus, applying the formula gives \( f = \frac{36}{2} = 18 \mathrm{~cm} \).
This calculation is crucial because the focal length is a fundamental part of the mirror equation. The focal length dictates how the mirror will converge or diverge light. Keep in mind:

• In a concave mirror, the focus is a point where parallel rays converge after reflection.
• It's important to remember that the focal length is always half of the radius for a spherical mirror.

Understanding these principles allows for better manipulation and solving of problems involving mirror image formation.

Determining Image Magnification

Magnification tells us how much larger or smaller an image is relative to the object. For mirrors, the magnification \( m \) is calculated using:
\[ m = -\frac{v}{u} \]In the exercise, \( v = 54 \mathrm{~cm} \) and \( u = -27 \mathrm{~cm} \). Thus, substituting these values gives:
\[ m = -\frac{54}{-27} = 2 \]This means the image is twice the size of the object and upright. Key things to note about magnification include:

• If \( |m| \) is greater than 1, the image is larger than the object.
• If \( m \) is positive, the image is upright. If negative, the image is inverted.
• Magnification helps assess whether more or less light reaches the eye, which influences how we perceive the image.

Expanding on this concept, if you move the object closer to the mirror, the magnification changes, thus affecting the size and nature of the image.