Question

When an small object is placed at a distance \(x_{1}\) and \(x_{2}\) from a lens on its principal axis, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then the focal length of the lens is : (1) \(x_{1}-x_{2}\) (2) \(\frac{x_{1}-x_{2}}{2}\) (3) \(\frac{x_{1}+x_{2}}{2}\) (4) \(x_{1}+x_{2}\) (5) \(\frac{x_{1}+x_{2}}{4}\)

Short answer

The focal length of the lens is \(\frac{x_1 + x_2}{2}\).

Step-by-step solution

Understand the problem

We're given an object placed at distances \(x_1\) and \(x_2\) from a lens, leading to a real image and a virtual image with the same transverse magnification. We need to find the focal length of the lens.

Recall lens formula for real and virtual images

The lens formula 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. For real images, \(v\) is positive, and for virtual images, \(v\) is negative.

Set up the equations using given distances

For the real image: \(u = -x_1\), for the virtual image: \(u = -x_2\). The object distances are negative because they are measured against the incident light direction.

Use magnification formulas

Transverse magnification \(m\) is given by \(m = \frac{-v}{u}\). Given \(m_1 = m_2\), the magnification for real image \(v_1\) and virtual image \(v_2\) must be equal.

Solve for the focal length using the lens formula

For the real image: \(\frac{1}{f} = \frac{1}{v_1} - \frac{1}{-x_1}\), solving gives: \(v_1 = \frac{f x_1}{x_1 - f}\). For the virtual image: \(\frac{1}{f} = \frac{1}{-v_2} - \frac{1}{-x_2}\), resulting in: \(v_2 = \frac{f x_2}{x_2 + f}\). Given \(\frac{-v_1}{-x_1} = \frac{v_2}{-x_2}\), multiply and simplify to find a relationship between \(x_1\) and \(x_2\).

Conclusion

Using the calculations and canceling terms appropriately, it can be shown that \(f = \frac{x_1 + x_2}{2}\). So, the focal length of the lens is \(\frac{x_1 + x_2}{2}\).

Key concepts

lens formula

The lens formula is fundamental in optics and is written as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f \) is the focal length of the lens, \( v \) is the distance from the lens to the image, and \( u \) is the distance from the object to the lens. This formula helps us understand how light rays converge or diverge when passing through a lens.

In the formula:

• If the image distance \( v \) is positive, the image formed is real and on the opposite side of the lens from the object.
• If \( v \) is negative, the image is virtual and on the same side of the lens as the object.

Using the lens formula can help you determine the focal length when object and image distances are known, which is crucial for solving many physics problems in optics.

transverse magnification

Transverse magnification \( m \) refers to how much larger or smaller the image is compared to the object, perpendicular to the lens axis. Mathematically, transverse magnification is given by \( m = \frac{-v}{u} \), where \( v \) is the image distance and \( u \) is the object distance.

Important points to note:

• A positive magnification indicates the image is upright relative to the object.
• A negative magnification indicates the image is inverted.
• For real images formed by a convex lens, magnification is typically negative, meaning the image is inverted.
• For virtual images, the magnification is positive, indicating that the image is upright.

By comparing the transverse magnification of real and virtual images, you can understand the nature and size of images formed by lenses.

real and virtual images

In optics, it's essential to distinguish between real and virtual images:

• Real Images: Formed when light rays converge at a point after passing through the lens. These images can be projected on a screen and are always inverted. They occur on the side of the lens opposite to the object.
• Virtual Images: Appear to come from a point on the same side of the lens as the object. These images cannot be projected on a screen as the light rays do not actually converge. They are always upright compared to the object.

Understanding these concepts aids in solving problems involving lenses, as it allows us to predict where and what kind of image a lens will form. Knowing whether an image is real or virtual also helps in determining the sign conventions used in the lens formula and magnification equations.