1No of Images will be calculated by (360/c)-1 where c is the angle between the two mirrors.
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
9hmm i know the formula but the derivation is tricky
ill be waiting for sunday ( b5555 ;) )
1
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...;)