Heya friends!!!!Pls Answer....
A 2cm high object is placed at a distance of 32cm from a concave mirror the image is real, inverted and 3cm in size. Find the focal length and position of image.
Pls solve it-_-_-_-_-&-_-_-_-_/
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Answered by
3
height of the object = 2cm
u = -32 cm
height of the image = -3cm
height of the image / height of the object = magnification
magnification = -3/2 = -v/u
3/2 = v/u
v = 3×-32/2 = -3×16 = -48cm
please mark it as the brainliest answer
u = -32 cm
height of the image = -3cm
height of the image / height of the object = magnification
magnification = -3/2 = -v/u
3/2 = v/u
v = 3×-32/2 = -3×16 = -48cm
please mark it as the brainliest answer
2Shashank1111:
Tell me why u use minus sigh on height of image...
it is because the image formed is inverted and the substance below the principal axis are taken as negative
In -v/u then what is the procedure
the magnification of the mirror has the formula -v/u
whatever be the sign , the answers will be the same...
OHK
Nyc
Answered by
5
Given :
Height of object ( O )= 2 cm
Height of image ( I ) = 3 cm
Distance of object ( u ) is 32 cm .
By formula :
-v / u = I / O
= > -v / 32 = 3 / 2
= > -v = 32 × 3 / 2
= > -v = 16 × 3
= > v =- 48
Hence image distance is 48 .
By lens formula :
1/f = 1/v - 1/u
Everything is negative in concave lens .
1/f = 1/48 - 1/32
= > 1/f = ( 2 - 3 ) / 96
= > 1/f = -1/96
f = - 96
Focal length is 96 cm.
Image is 48 cm in front of lens.
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