Question

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

Answer 1

Answer :

Step by step explanation :

h = 10 , f = 10 cm, u= - 15 cm

Using lens formula,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} = \frac{1}{30} \\ \\ v = 30 \: cm$

As v is +ve, image is real

m = v/u = h'/h

30/-15 = h'/2

h' = - 4

height of image is +ve which suggests that image is enlarged

-ve sign of magnification suggests that image is inverted.

Hence, the image is real, inverted and enlarged.

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Answer 2

$\bull$ Given :-

• Height of the object ( $h_o$ ) = 2 cm
• Focal length ( f ) = 10 cm
• Object distance ( u ) = - 15 cm

$\bull$ To Find :-

• The nature, position, and size of the image
• Magnification of the image

$\bull$ Concept :-

★ In this question, we will use the concept of branch of science known as "Light" under which we study, types of lenses and reflection and refraction respectively.

We will use two formulas for this question, they lens formula and Formula for magnification. They are as follows :-

✪ Lens Formula :-

$\bigstar {\color{orange}{\boxed{ \sf \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}}\bigstar$

Where,

• v = image distance
• u = object distance
• f = focal length

✪ Formula for Magnification :-

$\bigstar {\color {magenta} \boxed {\sf \: \frac{hi}{ho} = \frac{v}{u} }} \bigstar$

Where,

• hi = Height of image
• ho = Height of object
• v = image distance
• u = object distance

Now, we will simply put the given values and proceed further to obtain the required answers.

$\bull$ Solution :-

First of all, we will find the image distance.

$\sf \leadsto \: \frac{1}{10} = \frac{1}{v} - \frac{1}{ - 15} \: \: \: \: \\ \\ \sf \leadsto \: \frac{1}{v} = \frac{1}{10} + (\frac{1}{ - 15} ) \\ \\ \sf \leadsto \: \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \: \: \: \: \: \: \: \: \\ \\ \sf \leadsto \: \frac{1}{v} = \frac{3 - 2}{30} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \leadsto \: \frac{1}{v} = \frac{1}{30} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the value of v is 30 cm.

And since, v is greater than u, therefore, the image is enlarged.

Now, we will find the height of the image, therefore,

$\sf \leadsto \: \dfrac{hi}{2} = \dfrac{ 30}{ - 15} \: \: \: \: \: \: \: \\ \\ \sf \leadsto \: hi = \cancel\dfrac{30}{ - 15} \times 2 \\ \\ \sf \leadsto \: hi = - 2 \times 2 \: \: \: \\ \\ \sf \leadsto \: hi = - 4 \: \: \: \: \: \: \: \: \: \: \:$

And now, since height of image is in negative, therefore the image is inverted.

Hence,

• Nature of the object = Real and inverted
• Size = Enlarged

Now, we have to find it's magnification

Therefore,

$\sf \leadsto \: m = \frac{ - 4}{2} \\ \\ \sf \leadsto \: m = 2 \: \: \: \: \:$

Since, magnification can't be negative.

Hence, we have removed the negative sign.

✪ Answer :-

• Size of the image = Enlarged
• Nature of the image = Real and inverted
• Position = 30 cm away from the lens
• Magnification = 2