@VictoryGod forgot to make this thread
Bro I don’t think so
Cause ∆x = d Sin(theta). Me ye theta Change ho rha h… angle Phi se… Jisse Pattern ki Width and Intensity Change hongi… to obviously Pattern Waisa hi rahega…
Someone plz confirm
Haan agar 90° ho jaye to Circle ban sakte h…
Newton Ring k baare me suna ho to… [ Edit : Aaj hi interference padh rha tha to confuse ho gya )
Rotate to tb hona chahiye jab Source ki direction change ho jaye…( suppose Torch rakhi h ab usko rotate kiya… Idhr Slits ko rotate kr rhe)
Slits se light pass krni h…
bro but we are rotating in the same plane then D (distance b/w plane and slit) will still remain same…
fringe width me kya asar aayega?
spherical surface kaha h?
So sorry bro mera matlab isse tha…
Dekho jab phi 90° h to hyperbola se circle ban gya…
Rotation to slit ka kiya h idhr…
Pattern kyu rotate hoga??
Light to usi direction me jayegi jis direction me pehle jaa rhi thi
ek min…same plane me nhi ho rha rotation ???
zara type krke confirm krna toh @VictoryGod
Bro second line dekho… Last me “relative to original direction”
kisi se pucha bro?
my teacher was also confused
Mujhe diagram hi smjh nhi arha btaskte ho kaise arranged h ye slits? Smjh nhi arha
You take a sheet with a double slit and take another sheet having single slit (rotate it 90° to the original sheet) and superimpose it. The parts where slit is there on one sheet and plane on other sheet, so when superimposed they’ll give plane sheet on those parts.
If you closely observe after such superposition the arrangement will look like
two point sources and when the double slit is rotated then just the distance between the point sources will change (you can again imagine it that except those two points, remaining slits be covered when superimposed) and accordingly changes in the interference pattern we can see.
New distance between the point sources will be d sec(phi) now. @Vasu_Kashiv
A is incorrect… If sigma1>sigma2… then the fluid 2 will float on fluid 1
h2/h1 should be positive… sigma2>rho>sigma1
By drawing FBD for B, we find the acceleration of B to be g. Now imagine if B moves up by a distance x, the total slack created would be 2x, let the distance moved by A be y => distance moved by the pulley on left = y/2 => 2x = y - y/2 = y/2 => 4x = y. Differentiating both sides 2 times, we get acceleration of A to be 4×(acceleration of B)= 4g