Question

Que. 3) Answer the following.
1) 10cm high object is placed at a distance of 50em from a converging
focal length of 20cm determine the position, size and type of the image
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Answer 1

Explanation:

GIVEN:-

>> U = -10cm

>> f = 20cm

TO FIND:-

>> v and image's nature.

UNDERSTANDING THE CONCEPT:-

According to the question,

>>We know the value of U and f, So from this we can find the value of v by using lens formula.

$\huge\bf\underline{Answer:}$

>>According to Lens formula,

$\sf{ \dfrac{1}{v} - \dfrac{1}{u} \ = \dfrac{1}{f} }$

$\sf{ \dfrac{1}{v} - \dfrac{1}{10} = \dfrac{1}{20} }$

$\sf{v = - 20}$

Therefore, The value of v is -20.

Now magnification,

$\sf{ \dfrac{v}{2} = 2 }$

>>Hence image formed is virtual, erect and magnified two times.