Physics, asked by Anonymous, 2 months ago

\huge\sf\underline\orange{Question:-}

~ An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.​

Answers

Answered by IIMissCherryII
4

\bf\orange{♤Answer}

GIVEN THAT:

The height of object = 5cm

Position of object, u = – 25cm

The focal length of the lens, f = 10 cm

TO FIND:

The position of the image, v =?

Size of the image

Nature of the image

FORMULA:

We know that

1/v – 1/u = 1/f

Substituting the known values in the above equation we get,

1/v + 1/25 = 1/10

=> 1/v = 1/10 – 1/25

=> 1/v = (5 – 2)/50

Hence, 1/v = 3/50

So, v= 50/3 = 16.66 cm

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Now, we know that

Magnification = v/u

Hence, m = 16.66/-25 = -0.66

Also, we know that

m= height of image/height of the object

Or, -0.66 = height of image / 5 cm

Hence, height of image = -3.3 cm

The negative sign of the height of the image depicts that an inverted image is formed.

So, the position of image = At 16.66 cm on the opposite side of the lens

So, the position of image = At 16.66 cm on the opposite side of the lensSize of image = – 3.3 cm at the opposite side of the lens

So, the position of image = At 16.66 cm on the opposite side of the lensSize of image = – 3.3 cm at the opposite side of the lensNature of image – Real and inverted

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Answered by Anonymous
5

Given :-

  • Length of the object = 5cm
  • Position of the object = - 25 cm
  • Focal length = 10cm

To Find :-

  • Position of the image
  • Size of the image
  • Nature of the image

Formula used :-

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

Substituting the values we get :-

</p><p>  =  &gt; \frac{1}{v}  - ( -  \frac{1}{ 25} ) =  \frac{1}{10}  \\  =  &gt;  \frac{1}{v}  +  \frac{1}{25}  =  \frac{1}{10}  \:  \:  \:  \:  \:  \:  \:  \\ =  &gt;  \frac{1}{v}  =  \frac{1}{10}  -  \frac{1}{25}  \:  \:  \:  \:  \:  \:  \:  \\   =  &gt;  \frac{1}{v}  =  \frac{5 - 2}{50}  =   \frac{3}{50}  \\  =  &gt; v =  \frac{50}{3} = 16.66cm

We know that :-

  • Magnification = v/u

 =  &gt; Magnification =  \frac{16.66}{ - 25}  =  - 0.66

Also, we know that :-

 =  &gt; m =  \frac{height \: of \: the \: image}{height \: of \: object} </p><p> \\  =  &gt;  - 0.66 =  \frac{x}{5}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \: \\  =  &gt; x = 5 \times  - 0.66 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  =  &gt; x =  - 3.3cm \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

Here, the negative sign indicates that the image is inverted.

Therefore :-

  • Position of the image = 16.66 cm on the opposite side of the lens.

  • Size of the image = - 3.3 cm

  • Nature of the image = Real and inverted.

NOTE :- For the ray diagram kindly refer the attachment. :)

Hope the answer helps you mate :)

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