A convex lens of focal length $30$ cm is placed in contact with a concave lens of focal length $20$ cm. An object is placed at $20$ cm to the left of this lens system. The distance of the image from the lens in cm is ________ .
Options: A) $60/7$, B) $15$, C) $45$, D) $30$
Options: A) $60/7$, B) $15$, C) $45$, D) $30$
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Solution Steps
1
Assign Focal Lengths with Sign Convention
For a convex lens, $f_1 = +30$ cm.
For a concave lens, $f_2 = -20$ cm.
2
Calculate Equivalent Focal Length ($F$)
When lenses are in contact, the equivalent focal length $F$ is given by:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$
$$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = \frac{1}{30} – \frac{1}{20}$$
$$\frac{1}{F} = \frac{2 – 3}{60} = -\frac{1}{60}$$
So, $F = -60$ cm.
3
Apply Lens Formula
Given object distance $u = -20$ cm (to the left).
Using the lens formula: $$\frac{1}{v} – \frac{1}{u} = \frac{1}{F}$$
$$\frac{1}{v} – \frac{1}{-20} = \frac{1}{-60}$$
4
Calculate Image Distance ($v$)
$$\frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$$
$$\frac{1}{v} = -\frac{1}{60} – \frac{1}{20}$$
$$\frac{1}{v} = \frac{-1 – 3}{60} = -\frac{4}{60}$$
$$\frac{1}{v} = -\frac{1}{15}$$
Thus, $v = -15$ cm.
5
Interpretation of Result
The distance of the image from the lens is $15$ cm. The negative sign indicates that the image is virtual and formed on the same side as the object.
Final Answer: 15
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Theory
1. Combination of Thin Lenses in Contact
When two or more thin lenses are placed in contact with each other, they act as a single equivalent lens. The total power ($P$) of the combination is the algebraic sum of the individual powers: $P = P_1 + P_2 + \dots$. Since power is the reciprocal of focal length ($P = 1/f$), the equivalent focal length ($F$) is determined by the formula $1/F = 1/f_1 + 1/f_2 + \dots$. It is critical to use the correct sign convention (positive for converging/convex and negative for diverging/concave) when performing these calculations. This concept is widely used in optical instruments like telescopes and microscopes to minimize aberrations.
2. Lens Formula and Sign Convention
The lens formula, $1/v – 1/u = 1/f$, is the fundamental equation for determining image properties. According to the New Cartesian Sign Convention: 1) All distances are measured from the optical center. 2) Distances measured in the direction of incident light are positive. 3) Distances measured opposite to the direction of incident light (usually to the left) are negative. 4) Heights measured above the principal axis are positive, and below are negative. For a real object, $u$ is always negative. A positive $v$ indicates a real image on the opposite side, while a negative $v$ indicates a virtual image on the same side.
3. Nature of Converging and Diverging Systems
The nature of a lens system is determined by the sign of its equivalent focal length. If $F$ is positive, the system is converging (convex nature). If $F$ is negative, the system is diverging (concave nature). In this specific problem, although we combined a convex lens ($+30$ cm) and a concave lens ($-20$ cm), the resulting focal length was $-60$ cm. This means the diverging power of the concave lens outweighed the converging power of the convex lens, making the entire system behave as a single concave lens. This demonstrates how lens combinations can be tuned to achieve specific optical properties.
4. Magnification and Image Characteristics
Magnification ($m$) is defined as the ratio of the height of the image to the height of the object, which is also equal to $v/u$ for lenses. If $|m| > 1$, the image is magnified; if $|m| < 1$, it is diminished. A positive $m$ indicates an upright (virtual) image, while a negative $m$ indicates an inverted (real) image. In our problem, $m = v/u = (-15)/(-20) = +0.75$. This tells us the image is virtual, upright, and diminished to $75\%$ of the object's size. Such calculations are essential for designing camera lenses and corrective eyewear.
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FAQs
1
What if the lenses were separated by a distance?
If separated by distance $d$, the formula changes to $1/F = 1/f_1 + 1/f_2 – d/(f_1f_2)$.
2
Why is the concave lens focal length negative?
By convention, diverging lenses have a virtual principal focus, making their focal length negative.
3
Can the equivalent focal length be zero?
Only if $1/f_1 = -1/f_2$, which means the lenses have equal and opposite powers, effectively acting as a glass slab.
4
What does a negative image distance mean here?
It means the image is formed $15$ cm to the left of the lens system, making it virtual.
5
How is power related to this problem?
The powers of the lenses were $100/30 \approx +3.33$ D and $100/-20 = -5$ D. Net power is $-1.67$ D.
6
Does the order of lenses matter?
No, as long as they are thin and in contact, the equivalent focal length remains the same.
7
What is the magnification in this case?
Magnification $m = v/u = -15/-20 = 0.75$. The image is diminished.
8
Is the image real or virtual?
Since $v$ is negative (formed on the same side as the object), the image is virtual.
9
Why did we use cm instead of meters?
As long as all distances ($u, v, f$) are in the same units, the lens formula works correctly.
10
Can I use this for thick lenses?
No, these formulas are specific to thin lenses where the thickness is negligible compared to the focal lengths.
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