Problem 32 - Master Resource Book in JEE Main Physics

Question

A spherical mirror forms an image of magnification \(m=\pm 3 .\) The object distance, if focal length of mirror is \(24 \mathrm{~cm}\), may be (a) \(32 \mathrm{~cm}, 24 \mathrm{~cm}\) (b) \(32 \mathrm{~cm}, 16 \mathrm{~cm}\) (c) \(32 \mathrm{~cm}\) only (d) \(16 \mathrm{~cm}\) only

Step-by-Step Solution

Answer

Option (b) is correct: 32 cm and 16 cm.

Step 1: Understand the problem

We are given a spherical mirror with a focal length of \(24 \text{ cm}\) and a magnification of \(m = \pm 3\). We need to find the possible object distances from the given options.

Step 2: Use the magnification formula

The magnification \(m\) for mirrors is given by \(m = -\frac{v}{u}\), where \(v\) is the image distance and \(u\) is the object distance. We have \(m = \pm 3\). Hence, \(\pm 3 = -\frac{v}{u}\), leading to \(v = \mp 3u\).

Step 3: Apply the mirror equation

The mirror equation is \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). Given \(f = 24 \text{ cm}\) and \(v = \mp 3u\), substitute into the mirror equation:\[\frac{1}{24} = \frac{1}{\mp 3u} + \frac{1}{u}\]

Step 4: Solve for object distance (u)

Rearranging the equation, \[\frac{1}{24} = \frac{1}{u} \left( 1 \pm \frac{1}{3} \right)\]This simplifies to- For \(m = 3\), \(\frac{1}{24} = \frac{1}{u} \times \frac{4}{3}\), leading to \(u = 32 \text{ cm}\).- For \(m = -3\), \(\frac{1}{24} = \frac{1}{u} \times \frac{2}{3}\), leading to \(u = 16 \text{ cm}\).

Step 5: Match with given options

From the two object distances calculated, \(u = 32\text{ cm}\) and \(u = 16\text{ cm}\), compare with the options provided. Option (b) has both these distances: \(32\text{ cm}\) and \(16\text{ cm}\).

Key Concepts

Magnification

When we talk about magnification in optics, we're discussing how much larger or smaller an image appears compared to the actual object. Imagine looking through a magnifying glass or camera lens. This effect of magnification is captured by the magnification formula for mirrors, expressed as: \[ m = -\frac{v}{u} \] Here, \( m \) is the magnification, \( v \) is the image distance, and \( u \) is the object distance. In this formula:

• A positive magnification (\( m > 0 \)) indicates that the image is upright.
• A negative magnification (\( m < 0 \)) means the image is inverted.

In the exercise, with a magnification of \( \pm 3 \), it means the image can be either three times larger and inverted, or three times larger and upright, allowing multiple object distances depending on other properties like the focal length.

Spherical Mirror

A spherical mirror is a type of mirror that has a surface shaped like a section of a sphere. Such mirrors can be of two types:

• Concave mirrors: These curve inward, like a spoon.
• Convex mirrors: These curve outward.

Spherical mirrors are characterized by their ability to converge or diverge light sharply, creating unique image properties. The function of spherical mirrors forms part of the foundation of geometric optics. With these mirrors, critical formulas like the mirror equation and magnification relation help determine object positions, image positions, and characteristics, crucial in solving problems like the one given in the exercise.

Focal Length

The focal length \( f \) of a mirror is the distance between the mirror's surface and its focal point, where parallel rays of light either converge or appear to diverge from. This property depends on the mirror's curvature:

• For concave mirrors, the focal point is real and located in front of the mirror.
• For convex mirrors, the focal point is virtual and behind the mirror.

The core relationship for a spherical mirror is expressed by the mirror equation:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] With a given focal length, like \( 24 \text{ cm} \) in the exercise, this equation is used to figure out different object distances and image properties. By substituting known values into the formula, we can solve for missing distances or understand the mirror's function.

Object Distance

Object distance \( u \) is the distance from the mirror to the object. It is a crucial part of both the magnification and mirror equations. In geometrical optics, understanding object distance involves:

• Whether the object is within the focal length, at the focal point, or beyond.
• How this position affects the image size, orientation, and nature (real or virtual).

With given ranges and magnification constraints, object distance determines where an object must be placed for a specific image size and position. In the example problem, calculating possible object distances with a focal length of \( 24 \text{ cm} \) revealed two valid options. This demonstrates how the placement of an object can drastically change how, and where, the image appears.