Vedantu Notes for Lens Formula and Magnification - Lens Power
Lenses, both converging and diverging, are the marvels of optical physics that use the ability of these media to refract, reflect, or bend light rays. In general, the lenses come in two shapes: convex (curved outward) and concave (curved inward). One of their principal purposes is to magnify images, i.e., make images appear larger than their actual size. Nowadays, these lenses can be seen in microscopes, telescopes, binoculars, other optical instruments, and of course, in our own eyes. Scientists and students have many simple to complex algebraic equations to associate the shape and physical dimensions of a lens to the effects it puts on the light rays that pass through it. Here, we will learn and understand some of the most vital equations and formulae related to the lens, along with the lens power. We will also learn how to calculate magnification with the help of lens formula.
What is the Lens Formula?
Spherical lenses in optical physics are the lenses formed by coupling two spherical surfaces together. Based on this concept of formation by binding two surfaces, these lenses are of two types: convex lenses - the lenses formed by binding the two spherical surfaces curved outward and concave lenses - the lenses formed by binding the two spherical surfaces curved inward.
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Images formed by these lenses can be real, virtual, or of different sizes depending on the objects’ distance from the lens. Now, the Lens formula helps us in calculating the image distance. It is the formula, or we can say the equation that relates the focal length, the distance of the object, and the distance of the image for a lens. It is given as:
1/v + 1/o = 1/f
Where,
v = Distance of image formed from the optical center of the lens.
o = Distance of object from the optical center of the lens.
f = focal length of the lens.
The lens formula is applicable to both types of lenses - convex and concave. It can also be used to calculate image distance for both real and virtual images. If the equation provides a negative image distance, then the image formed is virtual and on the same side as the object. However, if the equation provides a negative focal length, then the lens is diverging, not converging.
Calculating Magnification By Using Lens Formula
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Magnification is defined as the ratio of the height of the image formed to the height of the object. In terms of distance of image and object, it is defined as the ratio of image distance to the object distance. For instance,
m = hv/ho = v/o
Where,
m = magnification
hv = height of the image
ho = height of the object
Example 1: If the distance of the object placed in front of a convex lens having a focal length of 10 cm is 15cm, find magnification. Also, tell the characteristics of the formed image.
Solution:
Focal length (f) = 10 cm (convex lens)
Object distance (o) = 15 cm
Putting the given values in 1/v+ 1/o = 1/f, we can get,
1/v = 1/f - 1/o
= 1/10 - 1/15
= 1/30
Therefore, v = image distance = 30 cm
Now, m = v/o
= 30/15
= 2 cm
From these results, we can say that the image is real, inverted, magnified 2 times, on the opposite side of the object, and at a distance of 30 cm from the lens.
Example 2: The distance of an object of height 6 cm from a concave lens is 20 cm. If its focal length is 10 cm, calculate the size and position of the image formed.
Solution:
As the lens here is concave, the focal length (f) = -10 cm
Object distance (o) = + 20 cm
Now by putting the values of (f) and (o) in 1/v+ 1/o = 1/f, we will get,
1/v = 1/f - 1/o
= - 1/10 - 1/20
= - 3/20
Therefore, image distance (v) = -20/3 = - 6.7 cm
Now, m = v/o
= (20/3)/20
= 1/3
And height of the image (hv) = ho x m
= 6 x 1/3
= 2 cm
From the results, we can conclude that the image is virtual, has a height of 2 cm, is on the same side as the object, and is at a distance of 6.7 cm from the concave lens.
Power of Lens
The power of a lens is its ability to converge the light rays falling on it. In other words, it is the measure of the degree of convergence or divergence of the rays of light falling on the lens. As the degree of convergence or divergence of the rays depends upon the focal length of the lens, the power of the lens can be defined as the reciprocal of the focal length of the lens. For instance, if the focal length (f) of a lens is 1 m, the power of the lens (p) is equal to 1/f = 1/1 = 1 dioptre. The SI unit of power of a lens is dioptre and often denoted by D. Note that as the focal length of a concave lens is negative, the power of this type of lens is negative (-), whereas the power of a convex lens is positive (+) as the focal length of this lens is positive.
Some common refraction phenomena include:
When a pencil is placed in a glass of water, it bends: When a pencil or stick is placed in a beaker or a glass of water, it seems slightly twisted. This occurs when light traveling from the rarer medium of air to the denser medium of water bends towards the incident, giving the impression of a bent pencil or stick.
Position of the fish in the pond's water: The ray from the pond's fish bends away from the incident's normal path. The emergent ray, which appears to be a fish, is seen just above its position.
Rainbow formation: After the rain, a rainbow appears. When a ray of light travels through raindrops, it is dispersed into its seven constituent colors, forming a rainbow in the sky.
Sun visibility slightly before sunrise: When the sun's rays enter the atmosphere (which is a denser material than vacuum), they bend away from normal to the incidence due to refraction. Because humans perceive the sun's refracted beams, the sun becomes visible shortly before sunrise.
FAQs on Lens Formula and Magnification
1. What is the Lens Formula, and how is it applied in CBSE Physics exams?
The Lens Formula relates the object distance (u), image distance (v), and focal length (f) of a lens: 1/v - 1/u = 1/f. In CBSE Physics exams, students are often asked to use this formula to calculate the position of an image formed by a convex or concave lens, typically providing two out of the three variables. Careful attention must be paid to sign conventions while applying this formula during calculations.
2. How is magnification defined for lenses, and what does a positive or negative value indicate?
Magnification in lenses is the ratio of the height of the image (h') to the height of the object (h), or alternatively, the ratio of image distance (v) to object distance (u): m = h'/h = v/u. A positive magnification signifies the image is virtual and erect, while a negative value shows the image is real and inverted. These distinctions are often probed in board exams for conceptual clarity.
3. Explain how the power of a lens is determined and mention its SI unit as per the latest syllabus.
The power of a lens (P) measures its ability to converge or diverge light rays and is given by the formula P = 1/f, where f is the focal length in meters. The SI unit of power is the dioptre (D). Positive power denotes a convex lens, and negative power indicates a concave lens, concepts directly aligned with board exam patterns.
4. Why is it important to follow sign conventions when solving lens formula questions in board papers?
Sign conventions are critical for determining the correct nature and position of images formed by lenses. In the lens formula, object distance (u) is typically negative if placed on the left of the lens, and focal length (f) is positive for convex and negative for concave lenses. Incorrect application of these conventions can lead to wrong answers, impacting scoring in CBSE Physics exams.
5. How are real and virtual images formed by convex and concave lenses assessed in important board questions?
In board questions, students may be asked to determine when a lens forms a real or virtual image. A convex lens forms a real, inverted image when the object is outside its focal point and a virtual, erect image when the object is within the focal length. Concave lenses always form a virtual, erect, and diminished image regardless of object position. These outcomes are essential for conceptual and application-based questions.
6. What types of numerical problems on lens formula and magnification are likely to appear in CBSE 2025–26 exams?
CBSE exams typically include numericals requiring calculation of image distance, object distance, magnification, height of image, or power of lens given appropriate data such as focal length and one distance. Marking schemes reward both correct calculation and application of the appropriate formula, with partial marks for correct set-up even if the answer is not fully solved.
7. Identify common mistakes students make while solving lens-related numerical questions and suggest exam strategies to avoid them.
Common errors include:
- Incorrect application of sign conventions
- Forgetting to convert focal length to meters when calculating power
- Mixing up object and image distances
- Ignoring the meaning of positive/negative magnification
8. How can understanding the formation of images through lenses help in real-life applications relevant for board exams?
Grasping how lenses form images aids in understanding devices like cameras, microscopes, telescopes, and corrective eyewear. Being able to link the physics of lenses to their practical use is often explored in board HOTS (Higher Order Thinking Skills) questions and strengthens conceptual knowledge for future studies or entrance exams.
9. What is the weightage of lens formula and magnification concepts in recent CBSE Physics board papers?
Typically, questions based on lens formula and magnification appear in the 3-mark or 5-mark sections of CBSE Physics papers. These may include direct numericals or application scenarios. Reviewing the latest sample paper pattern suggests at least one question from this area is highly probable in each exam cycle.
10. How should students approach revision for lens formula and magnification to maximize board exam performance?
Effective revision includes:
- Practicing a variety of numerical problems involving different lens types
- Summarizing key formulas and sign conventions
- Attempting previous years’ exam questions
- Understanding image formation scenarios through ray diagrams
