9hmm i know the formula but the derivation is tricky
ill be waiting for sunday ( b5555 ;) )
try doing this if u ve free time
1)
two mirrors are inclined at an angle θ
find the no of images formed
2)
a person enters a room with ceiling and 2 adjacent walls as mirrors .
how many images can he see ?
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28 Answers
91No of Images will be calculated by (360/c)-1 where c is the angle between the two mirrors.
1the first one can be analysed by observing that betweeen any three images angle 2θ between any five angle 4θ ... ultimately when they coincide we get 2π/θ but the object is to be not considered so we get 2π/θ - 1
13 images each of d wall n ceilin are visible.............. rit???
1answer for 1st is a standard one...'
for second one it is 2, though 3 images r formed only 2 are visible
62I have not known a single book which gives the correct explanation for part 1 of this question! :)
But this is worth thinking in free time only :)
33lol..
where it is fraction...
theres [] hai naa...gr8est integer function...
1ab dekho, mujhe kabhi phy6 aur math mein logic saath nahi deta...
but how can no. of images be a fraction?
33But it will be better to do from the basic because itna yaad nahi rahta :(
331st one..
case 1.
If 360/θ is even...
then no of images= 360/θ-1
case 2.
If 360/θ is odd
2.1 object at angle bisector
then no of images=360/θ
2.2 object not at angle bisector
no of images =360/θ-1
case 3.
if 360/θ is fraction then
no of images [360/θ]
39for 1st one if no one gives explanation by sunday, i will. i have a beautiful explanation, trust me.
132nd one
infinite numbers of images theoretically
but a few practically
39late post tha............. by the way, abhirup formula nahiin try to derive it, and ya its only for 11th waale.
39@ritika, it does not depend on the placement of object { in a way depends but not here]. u only have to...........
1wer's d object placed in the 1st question?
Nd i'll assume d person is luminiscent...;)