Physics, asked by prashantjaiswal1, 1 year ago

Answer please (see question In pic) 50 pts.

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Answered by Prashant24IITBHU

image distance of near end
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{30}$

$\frac{1}{v} = \frac{2}{30}$

$v= 15 cm$

similarly distance of image of rear end from thr mirror
$v= \frac{40}{3} cm$

So, lenght of the base of the wire = 15-40/3 = 5/3cm

Prashant24IITBHU: Magnification for the far end = (-v/u) = (-5/(3x40)) =(-1/24). So lenght of far end will be (10/24) = (5/12) cm

prashantjaiswal1: just see the options and reply me the correct answer

prashantjaiswal1: please please

prashantjaiswal1: reply

Prashant24IITBHU: 10 is the right answer

prashantjaiswal1: how

Prashant24IITBHU: (10/3)+(5/3)+5 = 10

Prashant24IITBHU: magnification of the near end =(-v/u) =(-40/(3x40)) =(-1/3). So, lenght of the near end = (10/3)

Prashant24IITBHU: sorry this was lenght of the far end

Prashant24IITBHU: So,Total lenght = Lenght of the near end + lenght of the base + lenght of the far end = (5)+(5/3)+(10/3) = 10

Answered by itzjaishankar

Answer:

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