24 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.
where am i going wrong
@integralofe^2v
@vj25_ yeh wala ?
v aur u ka mu exchange nahi hoga?matlab 1/v-1.33/u
in case 2 ?
why ?
u3/v - u1/u = u2-u1/r1 + u3-u2/r2
aisa hi toh tha
o accha right se lia hai mb
ek cheez samajh nahi aai
tune primary aur secondary focus nikala hai right
yep
ek baari ray glass mein enter karegi
aur ek baari exit
system ka net focal length nikalne ke lie dono ko alag alag nikala
not sure about this method ek bar kal khud nikal ke dekhunga
lag toh sahi rha hai
alrr
can we consider the ext medium for the concavo convex as water though? what abt the air water interface.. i mean it's fine for normal incidence but otherwise it wont be right?
For the concave portion of mirror ext med is water only
There's no air between the water and the concave part
What formula did u even use here i cant make it out😠i just used the lens maker one and got the ans
yup sorry i was thinkin of smtg else.. power ko vese phir mu/feq likhna pdega na
like for first thing u calculated since mu3 is 1 it'll be 1/f = 1/45.. 2nd case matlab 2nd refraction mu3 is now 4/3.. so 4/3*60 = 1/45 ... 1/45+1/45+1/10= 13/90
ig should give 90/13
bcuz in these lenses power is mu3/f
my main doubt is focal length of lens should be diff when ray are coming from right as compared to left
thats not the ans
i hv used mirror formula only
What's the ans for focal length? F= 90/13 so 2F = 180/13.. so you'll have to subtract according to whatvr u get from part 1
You can't write 1/f as power for the second refraction here.
Not the final ans. Value of focal length
So over here where you've taken 1/60 as 1/f2 take 4/3*60 = 1/45
huh why
yeah u = 180/13
par kyu
ugh ill recheck in a while
bhai close mt krio
power of curved surface lagana hai
bcuz of the medium it'll change
where did u learn this
it's taught generally... a pic from cengage if u want for ref
and another thing.. even though focal lengths depend on the direction of incidence power doesnt... so left to right, right to left will be same always