PLEASE ANSWER FOR THIS PLEASE
Answers
Answer:
Mathematics includes the study of such topics as quantity, structure, space, and change. It has no generally accepted definition. Mathematicians seek and use patterns to formulate new conjectures; they resolve the truth or falsity of such by mathematical proof.
Answer:
⚘ Question :-
An object 2 cm height is placed 20 cm in front of convex lens of focal length 10 cm. Find a distance from the lens at which a screen should be placed in order to obtained a sharp image. What will be size and nature of image formed?
⚘ Answer :-
Distance from the lens at which a screen should be placed is 20 cm.
Size of image is 2 cm.
Nature of image is real, inverted.
Explanation:
⚘ Given :-
Height of object (h) = 2 cm
Distance of object (u) = -20 cm
Focal length of lens (f) = +10 cm
⚘ To Find :-
Distance from lens at which screen should be placed (v)?
Size of image (h')?
Nature of image?
⚘ Solution :-
Firstly let's calculate distance from lens at which screen should be placed (v) by using len's formula ::
We know that,
\leadsto{\large{\boxed{\sf{\red{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}}}}⇝
v
1
−
u
1
=
f
1
Where,
v is distance of screen
u is distance of object
f is focal length of lens
We have,
u = -20 cm
f = 10 cm
v = ?
According to the question by using the formula we get,
➨ \sf \dfrac{1}{v} - \dfrac{1}{-20} = \dfrac{1}{10}
v
1
−
−20
1
=
10
1
➨ \sf \dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{-20}
v
1
=
10
1
+
−20
1
➨ \sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20}
v
1
=
10
1
−
20
1
➨ \sf \dfrac{1}{v} = \dfrac{2 - 1}{20}
v
1
=
20
2−1
➨ \sf \dfrac{1}{v} = \dfrac{1}{20}
v
1
=
20
1
By doing cross multiplication we get,
➨ \bf\purple{v = 20}v=20
∴ Distance from lens at which screen should be placed is 20 cm.
Now, let's calculate size of the image (h') by using magnification formula ::
We know that,
\leadsto{\large{\boxed{\sf{\blue{m = \dfrac{v}{u} = \dfrac{h'}{h}}}}}}⇝
m=
u
v
=
h
h
′
We have,
v = 20 cm
u = -20 cm
h = 2 cm
h' = ?
According to the question by using the formula we get,
➨ \sf \dfrac{v}{u} = \dfrac{h'}{h}
u
v
=
h
h
′
➨ \sf {\cancel{\dfrac{20}{-20}}} = \dfrac{h'}{2}
−20
20
=
2
h
′
➨ \sf \dfrac{1}{-1} = \dfrac{h'}{2}
−1
1
=
2
h
′
➨ \sf -1 = \dfrac{h'}{2}−1=
2
h
′
➨ \sf h' = -1\:\times\:2h
′
=−1×2
➨ \bf\pink{h' = -2}h
′
=−2
∴ Size of image is 2 cm.
Now,
As sign is negative therefore image is real, inverted