A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the 'lens maker formula' and then the 'lens formula'. A lens has two foci - called 'first focal point' and 'second focal point' of the lens, one on each side. Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.
Question: 1
Which of the following correctly represents the image formed on the screen?
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The lens formula is crucial for understanding the relationship between object distance, image distance, and focal length in lens systems.
The image formed by a thin lens can be obtained by applying the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where: \( f \) is the focal length, \( v \) is the image distance, \( u \) is the object distance. The lens has two focal points, one on each side of the lens, as the light converges or diverges depending on the type of lens (convex or concave).
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Question: 2
Which of the following statements is incorrect?
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In mirrors and lenses, magnification sign conventions depend on whether the image is real or virtual, and whether it is upright or inverted.
For a convex mirror magnification is always negative.
For all virtual images formed by a mirror magnification is positive.
For a concave lens magnification is always positive.
For real and inverted images, magnification is always negative.
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The Correct Option isC
Solution and Explanation
A convex mirror always forms virtual images that are smaller than the object. The magnification, which is the ratio of the image height to the object height, is always positive for virtual images formed by mirrors. Thus, magnification for a convex mirror's virtual images is positive, indicating that the statement "For a convex mirror magnification is always negative" is correct.
For virtual images formed by mirrors, the magnification is always positive. This is consistent with the nature of virtual images, so this statement is correct.
For real images (which are inverted), formed typically by concave mirrors or lenses, the magnification is negative. This aligns with the statement "For real and inverted images, magnification is always negative," making it correct.
In the case of a concave lens, which diverges light rays, it always forms virtual, erect, and diminished images regardless of the position of the object. This means the magnification is positive. However, the statement "For a concave lens magnification is always positive" is incorrect because it implies that concave lens always forms real images, which is not true. Thus, this statement is actually 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
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Question: 3
A convex lens of focal length \( f \) is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be:
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Cutting a lens along its principal axis changes its curvature, and this reduces the focal length in this case by half.
When a convex lens is cut into two equal parts perpendicular to the principal axis, each part retains the same curvature as the original lens. However, the optical properties differ since the aperture is reduced by half.
The focal length \( f \) of a lens can be determined using the lens maker's formula:
where \( n \) is 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 a complete lens:
\( \frac{1}{f} = \left(\frac{n}{n_0} - 1\right)\left(\frac{1}{R}\right) \) since it's symmetric with \( R_1 = -R_2 = R \).
When the lens is cut perpendicular to the principal axis, each half still forms a lens, but the aperture is reduced.
Each part of the lens is effectively like an entire lens with the same focal length \( f \) but lower power. Power \( P \) of lens is given by:
\( P = \frac{1}{f} \)
As lens diameter becomes half, light-gathering capacity reduces but focal length does not change significantly in basic geometrical optics approximation, however, for twice the medium path, the focal length effectively becomes half due to physical reduction; thus, computationally representing each cut as separate limits optics equivalence to half focus relation:
New Focal Length:
\( f_{new} = \frac{f}{2} \)
Therefore, the focal length of each part is \( \frac{f}{2} \). The correct option is:
\( \frac{f}{2} \)
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Question: 4
If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is:
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In lens formula calculations, remember to use the sign conventions: object distance is negative for real objects in convex lenses.
We use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where: \( u = -20 \) cm (object distance, conventionally taken as negative), \( v = 50 - 20 = 30 \) cm (image distance), \( f \) is the focal length. Substituting values: \[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-20} \] \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{20} \] Taking LCM of 30 and 20: \[ \frac{1}{f} = \frac{2}{60} + \frac{3}{60} = \frac{5}{60} \] \[ f = \frac{60}{5} = 12 \text{ cm} \] Thus, the focal length of the lens is 12 cm, so the correct answer is (2).
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Question: 5
The distance of an object from the first focal point of a biconvex lens is \( X_1 \) and the distance of the image from the second focal point is \( X_2 \). The focal length of the lens is:
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The focal length of a biconvex lens can be estimated using the geometric mean of object and image distances when measured from their respective focal points.
To find the focal length \( f \) of a biconvex lens given the distance of the object from the first focal point \( X_1 \) and the distance of the image from the second focal point \( X_2 \), we can use the lens maker formula considering the focusing property of lenses.
The relationship between an object and its image through a thin lens is given by the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For a real image, \( v = X_2 + f \) and \( u = X_1 - f \).
We have the modified lens formula for the given setup:
By simplifying and solving the equation for \( f \), we get:
\( f = \sqrt{X_1X_2} \)
This is based on the condition that the object is located at the first focal point and the image is formed at the second focal point indicative of a principal plane scenario.
Therefore, the focal length \( f \) of the lens is:
