Physics, asked by hydratejes, 9 months ago

A 2.0cm tall object is placed perpendicular to the principal axix of a convex lens of focal length 10cm . The distance of the object from the lens is 15 cm . Find the nature, position and size of the image . Also find its magnification.​

Answers

Answered by Atαrαh
3

We know that ,

focal length of a convex lens is positive

f=+10cm

the object distance Is also negative

u=-15cm

As per the lens formula,

 \frac{1}{f}  =  \frac{1}{v}  - \frac{1}{u}  \\  \frac{1}{v}  =   \frac{ + 1}{10}   -\frac{1}{15}   =  \frac{ - 5}{150}   \\ v =   30cm

m =  \frac{  v}{u}  =  \frac{hi}{ho}

hi =  \frac{30 \times 2}{ - 15} =  - 4cm

the image is inverted , enlarged and real

I hope this helps ( ╹▽╹ )

Answered by Anonymous
2

Answer:

Height of the object $h _{ 0 }$ = +2.0 cm

Focal length f = +10 cm

Object-distance u = -15cm

Image-distance v = ?

1/v - 1/u = 1/f ====>1/v = 1/f + 1/u ===> 1/v = 1/10 + 1/-15

1/v = -2+3/30

1/v = 1/30

v = 30 cm

The positive sign of v shows that the image is formed at a distance of 30 cm on the other side of the optical centre.The image is real and inverted.

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