KCET Electromagnetic Waves Weightage 2025

KCET Electromagnetic Waves Weightage 2025: Electromagnetic Waves is one of the important chapters for the Physics section in the KCET 2025 exam. KCET Electromagnetic Waves Weightage 2025 is expected to be 2% of the total Physics Section.  Students planning to appear for the KCET exam can expect 2 to 3 questions on this topic. Displacement Current, Inconsistency of Ampere’s Circuital Law, Ampere-Maxwell Law, Electromagnetic Waves, Electromagnetic Spectrum and Electromagnetic Characteristics are some of the important topics of Electromagnetic Waves. To score in the KCET 2025 exam , it is beneficial to be aware of the KCET syllabus 2025 , and their weightage so that they can plan their study schedule effectively.

Latest update: KCET 2025 exam dates have been officially announced. KCET 2025 exam will be conducted on April 16 and 17, 2025.

Expected KCET Electromagnetic Waves Weightage 2025

Also Check - KCET Subject Wise Syllabus 2025

KCET Electromagnetic Waves Weightage of Last 5 Years

Knowing the weight of each head is key to doing well on any test. KCET Electromagnetic Waves Weightage of the last 5 years has been provided in the following table.

With this kind of weightage, it is clear that this topic is important, and understanding its basic and advanced concepts will greatly aid in scoring well on the exam.

Important Topics of Electromagnetic Waves for KCET 2025

Electromagnetic waves are among the concepts students should be familiar with. Below are the key topics to focus on to score perfect marks in questions on electromagnetic waves:

Note: This table will help you know the weightage of each topic’s information and weightage in the Electromagnetic waves chapter.

Sample Questions on Electromagnetic Waves for KCET 2025

Here are some sample questions that will appear in the KCET 2025 exam. Addressing these will help you get an idea of ​​the types of questions you may encounter:

1. What is the wavelength of an electromagnetic wave with a frequency of 5×1014 Hz5 \times 10^{14} \, \text{Hz}5×1014Hz in a vacuum?

Solution:
The wavelength λ\lambdaλ is given by the formula:

λ=cf\lambda = \frac{c}{f}λ=fc​

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s and f=5×1014 Hzf = 5 \times 10^{14} \, \text{Hz}f=5×1014Hz.

λ=3×1085×1014=6×10−7 m=600 nm\lambda = \frac{3 \times 10^8}{5 \times 10^{14}} = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}λ=5×10143×108​=6×10−7m=600nm

So, the wavelength is 600 nm.

2. Calculate the speed of electromagnetic waves in a material with an index of refraction n=2n = 2n=2.

Solution:
The speed vvv in a material is calculated as:

v=cnv = \frac{c}{n}v=nc​

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s and n=2n = 2n=2.

v=3×1082=1.5×108 m/sv = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \, \text{m/s}v=23×108​=1.5×108m/s

The speed of the wave in this material is 1.5×108 m/s1.5 \times 10^8 \, \text{m/s}1.5×108m/s.

3. What is the frequency of an electromagnetic wave with a wavelength of 400 nm400 \, \text{nm}400nm in a vacuum?

Solution:
The frequency fff can be found using:

f=cλf = \frac{c}{\lambda}f=λc​

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s and λ=400 nm=400×10−9 m\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}λ=400nm=400×10−9m.

f=3×108400×10−9=7.5×1014 Hzf = \frac{3 \times 10^8}{400 \times 10^{-9}} = 7.5 \times 10^{14} \, \text{Hz}f=400×10−93×108​=7.5×1014Hz

So, the frequency is 7.5×1014 Hz7.5 \times 10^{14} \, \text{Hz}7.5×1014Hz.

4. Explain the concept of polarization in electromagnetic waves.

Solution:
Polarization refers to the orientation of the electric field vector in an electromagnetic wave. In unpolarized light, the electric field oscillates in multiple directions perpendicular to the direction of wave propagation. In polarized light, the electric field oscillates in a single direction. Polarization is commonly used in sunglasses and camera lenses to reduce glare by filtering certain orientations of light waves.

5. A light wave travels from air (n = 1) into water (n = 1.33). Calculate its speed in water.

Solution:
The speed vvv of light in water is:

v=cnv = \frac{c}{n}v=nc​

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s and n=1.33n = 1.33n=1.33.

v=3×1081.33≈2.26×108 m/sv = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \, \text{m/s}v=1.333×108​≈2.26×108m/s

The speed of light in water is approximately 2.26×108 m/s2.26 \times 10^8 \, \text{m/s}2.26×108m/s.

6. If the wavelength of a wave doubles, what happens to the frequency if the speed remains constant?

Solution:
Using the relation c=fλc = f \lambdac=fλ, if the wavelength λ\lambdaλ doubles and ccc remains constant, the frequency fff must halve.

So, if the wavelength doubles, the frequency will halve.

7. A ray of light with a wavelength of 500 nm500 \, \text{nm}500nm in air enters a glass medium with an index of refraction n=1.5n = 1.5n=1.5. What is the wavelength of the light inside the glass?

Solution:
The wavelength in the glass λglass\lambda_{\text{glass}}λglass​ is calculated using:

λglass=λairn\lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n}λglass​=nλair​​

where λair=500 nm\lambda_{\text{air}} = 500 \, \text{nm}λair​=500nm and n=1.5n = 1.5n=1.5.

λglass=5001.5≈333 nm\lambda_{\text{glass}} = \frac{500}{1.5} \approx 333 \, \text{nm}λglass​=1.5500​≈333nm

The wavelength of the light inside the glass is approximately 333 nm.

Practice Questions for KCET 2025:

Conclusion

The topic of Electromagnetic Waves is an important part of the KCET Physics syllabus , carrying a weightage of 2%. By understanding the basic principles, applying mathematical problems, and reviewing key concepts regularly, you will be able to master this and improve your overall performance in the KCET 2025 exam.

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