## Sunday, January 23, 2011

### Communication Systems -Two Questions Involving Line of Sight Space Wave Communication

Questions on communication systems at Class 12 level are often simple but occasionally you may get questions which are a little bit confusing and time consuming. I give below a couple of practice questions for you:

(1) A television transmitting antenna is mounted at a height of 100 m. For satisfactory line of sight communication at a distance of 40 km, what should be the minimum height of the receiving antenna? (Radius of the earth = 6400 km).

(a) 20 m

(b) 25 m

(c) 30 m

(d) 35 m

(e) 40 m

If ht and hr represent the heights of the transmitting antenna and the receiving antenna respectively, the maximum separation (d) between them for satisfactory line of sight communication is given by

d = √(2Rht) + √(2Rhr) where R is the radius of the earth.

Therefore, we have

40 = √(2×6400×0.1) + √(2×6400×hr)

[Note that we have expressed all distances in kilometre and hence we will obtain the height of the receiving antenna in km].

Squaring, 1600 = 2×6400 (0.1 + hr)

Therefore, (0.1 + hr) = 1/8 from which

hr = 0.025 km = 25 m.

(2) The minimum service area covered by a TV transmitter antenna mounted at a height h is (radius of the earth = R)

(a) πR2h

(b) 2πR2h

(c) πh2R

(d) πRh

(e) Rh

The coverage area will be minimum when the receiving antenna is at the ground level. In other words, the height of the receiving antenna is zero.

The distance (d) up to which line of sight communication is possible is therefore given by

d = √(2Rht) where ht is the height of the transmitting antenna.

The minimum coverage area is πR2d =π[√(2Rht)]2 = Rht = Rh

You can access all posts on communication systems on this site by clicking on the label, ’communication system below this post.

## Monday, January 17, 2011

### Apply for Entrance Examination for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2011 (KEAM 2011), Kerala

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2011-12.

(a) Engineering: B.Tech, B.Tech. (Agricultural Engg.) and B.Tech. (Dairy Sc. & Tech.)

(b) Architecture: B.Arch.

(c) Medical: (i) MBBS (ii) BDS (iii) BAMS (iv) BSMS (v) BHMS

(d) Agriculture and Allied Courses: (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(e) Veterinary: BVSc. & AH

Dates of Exam:

Engineering Entrance Examination

18.04.2011 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

19.04.2011 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

20.04.2011 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

20.04.2011 2 P.M. to 4.30 P.M. Paper-II: Biology.

Application form and Prospectus will be distributed from 19.1.2011(Wednesday) to 14.02.2011 (Monday) through selected Post Offices.

Last Date for receipt of Application by CEE: 14-02-2011 (Monday) - 5 PM

You will find complete details and information updates at http://www.cee-kerala.org/

* * * * * * * * * * * * * * *

To access earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the search button.

## Friday, January 14, 2011

### Publishing now on Custom Domain

Some of you might have noted that when you search for a page in ‘physicsplus’ at the URL with domain name physicsplus.blogspot.com, you get the required page at physicsplus.in. Please note that I have started publishing using custom domain name physicsplus.in.

So the blog you find is the same and no need to worry!

## Thursday, January 06, 2011

### Two Questions (MCQ) on Photo Diodes

“He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.”

– Albert Einstein

As you know, photo diodes are special purpose p-n junction diodes. They are very popular photo detectors used for detecting optical signals in communication systems.

Today we will discuss two multiple choice questions on photo diodes:

(1) In using a photo diode as a photo detector, it is invariably reverse biased. Why?

(a) The power consumption is much reduced compared to reverse biased condition

(b) Electron hole pairs can be produced by the incident photons only if the photo diode is reverse biased

(c) Light variations can be converted into current variations only if the photo diode is reverse biased

(d) When photons are incident on the diode, the fractional change in the reverse current is much greater than the fractional change in the forward current

(e) The photo diode will be spoilt if it is operated under forward biased condition

Whether the photo diode is reverse biased or forward biased, the number of electron hole pairs produced by the incident photons is the same. In other words, the change in the diode current is the same in both cases. But in the reverse biased condition the current drawn by the diode in the absence of the photons is extremely small, of the order of nanoamperes or microamperes where as in the forward biased condition this current is significant, of the order of tens of milliamperes. The fractional change in the current because of the incident photons is therefore large and easily measurable if the photo diode is reverse biased. The correct option is (d).

(2) The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about

(a) 620 nm

(b) 700 nm

(c) 740 nm

(d) 860 nm

(e)1240 nm

The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by

λE = 12400, very nearly.

Therefore, λ = 12400/E

[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].

Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.

Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (a).

## Saturday, January 01, 2011

“Cheers to a New Year and another chance for us to get it right”

- Oprah Winfrey

“Happy New Year 2011”