Question

Which of the following correctly represents the image formed on the screen?

• A.
• B.
• C.
• D.
• A. For a convex mirror magnification is always negative.
• B. For all virtual images formed by a mirror magnification is positive.
• C. For a concave lens magnification is always positive.
• D. For real and inverted images, magnification is always negative.
• A. \( f \)
• B. \( 2f \)
• C. \( \frac{f}{2} \)
• D. \( \frac{f}{4} \)
• A. 10 cm
• B. 12 cm
• C. 16 cm
• D. 20 cm
• A. \( X_1 X_2 \)
• B. \( \sqrt{X_1 + X_2} \)
• C. \( \sqrt{X_1 X_2} \)
• D. \( \sqrt{\frac{X_2}{X_1}} \)

Answer 1

The Correct Option is B

The lens formula, \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \], determines the image position. Here, \( f \) represents focal length, \( v \) is image distance, and \( u \) is object distance. Lenses possess two focal points, located on opposing sides, influencing light convergence or divergence based on whether the lens is convex or concave.

Answer 2

A convex mirror consistently produces virtual images that are smaller than the object. Magnification, defined as the ratio of image height to object height, is always positive for virtual images formed by mirrors. Therefore, the magnification for a convex mirror's virtual images is positive, confirming that the statement "For a convex mirror magnification is always negative" is correct.

For virtual images formed by mirrors, magnification is invariably positive, aligning with the nature of virtual images. Consequently, this statement is correct.

Real images, which are inverted, typically formed by concave mirrors or lenses, exhibit negative magnification. This corresponds to the statement "For real and inverted images, magnification is always negative," rendering it correct.

A concave lens, which diverges light, invariably forms virtual, erect, and diminished images irrespective of object placement. This results in positive magnification. However, the statement "For a concave lens magnification is always positive" is inaccurate as it implies the formation of real images, which is not universally true for concave lenses. Thus, this statement is incorrect.

Statement	Correctness
For a convex mirror magnification is always negative.	Correct
For all virtual images formed by a mirror magnification is positive.	Correct
For a concave lens magnification is always positive.	Incorrect
For real and inverted images magnification is always negative.	Correct

Answer 3

When a convex lens is divided into two equal halves perpendicular to its principal axis, each resultant part maintains the original curvature. However, their optical performance is altered due to a halved aperture.

The focal length \( f \) of a lens is determined by the lens maker's formula:

\( \frac{1}{f} = \left(\frac{n}{n_0} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)

Here, \( n \) represents the refractive index of the lens material, \( n_0 \) is the refractive index of the surrounding medium, and \( R_1 \), \( R_2 \) are the radii of curvature of the lens surfaces.

For an intact, symmetric lens where \( R_1 = -R_2 = R \), the formula simplifies to:

\( \frac{1}{f} = \left(\frac{n}{n_0} - 1\right)\left(\frac{1}{R}\right) \)

Upon cutting the lens perpendicular to the principal axis, each fragment functions as a lens with a reduced aperture.

Each half acts as a complete lens with the same focal length \( f \), but its power is diminished. Lens power \( P \) is defined as:

\( P = \frac{1}{f} \)

While the reduction in lens diameter halves its light-gathering capacity, the focal length remains largely unchanged in basic geometrical optics. However, considering the doubled path length of light through the medium within each half, the effective focal length becomes half. This physical reduction implies a new focal length calculation:

New Focal Length:

\( f_{new} = \frac{f}{2} \)

Consequently, the focal length of each individual part is \( \frac{f}{2} \). The correct option is:

\( \frac{f}{2} \)

Answer 4

Applying the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]. Given: \( u = -20 \) cm (object distance, negative by convention), \( v = 50 - 20 = 30 \) cm (image distance). \( f \) represents the focal length. Substituting the values into the formula yields: \[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-20} \] which simplifies to \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{20} \]. To find the common denominator for 30 and 20 (LCM is 60), we get: \[ \frac{1}{f} = \frac{2}{60} + \frac{3}{60} = \frac{5}{60} \]. Solving for \( f \): \[ f = \frac{60}{5} = 12 \text{ cm} \]. Therefore, the focal length of the lens is 12 cm, corresponding to option (2).

Answer 5

The focal length \( f \) of a biconvex lens, given the distance of the object from the first focal point as \( X_1 \) and the distance of the image from the second focal point as \( X_2 \), can be determined using the lens maker formula and the lens formula.

The general 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 a real image formation, the image distance from the lens is \( v = X_2 + f \) and the object distance from the lens is \( u = X_1 - f \).

Substituting these into the lens formula yields the modified equation for this specific configuration:

\( \frac{1}{f} = \frac{1}{X_2+f} + \frac{1}{X_1-f} \)

Simplifying and solving this equation for \( f \) results in:

\( f = \sqrt{X_1X_2} \)

This formula is derived under the assumption that the object is positioned at the first focal point and the image is formed at the second focal point, characteristic of a principal plane scenario.

Consequently, the focal length \( f \) of the lens is:

\( \sqrt{X_1 X_2} \)