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Answers
Answer:
concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens ? Also calculate the size of the image formed .
Solution :
Focal length of the concave lens (f) = - 20 cm
Object distance of the concave lens (u) = ? cm
Image distance of the concave lens (v) = - 15 cm
Height of the object (H₀) = 5 cm
Height of the image (Hᵢ) = ? cm
Apply Lens formula ,
\longrightarrow \sf \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{f}}}=\dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{v}}}-\dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{u}}}⟶
f
1
=
v
1
−
u
1
\longrightarrow \sf \dfrac{1}{-20}=\dfrac{1}{-15}-\dfrac{1}{u}⟶
−20
1
=
−15
1
−
u
1
\longrightarrow \sf \dfrac{1}{u}=\dfrac{1}{20}-\dfrac{1}{15}⟶
u
1
=
20
1
−
15
1
\longrightarrow \sf \dfrac{1}{u}=\dfrac{3-4}{60}⟶
u
1
=
60
3−4
\longrightarrow \sf \dfrac{1}{u}=\dfrac{-1}{60}⟶
u
1
=
60
−1
\longrightarrow \sf \dfrac{1}{u}=\dfrac{1}{-60}⟶
u
1
=
−60
1
\longrightarrow \textsf{\textbf{u }} \textsf{\textbf{= }} \textsf{\textbf{- 60 cm}}⟶u = - 60 cm
Object should be placed at 60 cm from left side of the lens .
Apply magnification ,
\Longrightarrow \sf \textsf{\textbf{m}} \textsf{\textbf{ = }}\dfrac{\textsf{\textbf{v}}}{\textsf{\textbf{u}}}⟹m =
u
v
\Longrightarrow \sf \dfrac{H_i}{H_o}=\dfrac{v}{u}⟹
H
o
H
i
=
u
v
\Longrightarrow \sf \dfrac{H_i}{5}=\dfrac{-15}{-60}⟹
5
H
i
=
−60
−15
\Longrightarrow \sf \dfrac{H_i}{5}=\dfrac{1}{4}⟹
5
H
i
=
4
1
\Longrightarrow \sf H_i=\dfrac{5}{4}\ cm⟹H
i
=
4
5
cm
\Longrightarrow {\textsf{\textbf{H}}}_{\textsf{\textbf{i}}} \textsf{\textbf{ = }} \textsf{\textbf{1.25 cm}}⟹H
i
= 1.25 cm
Height/Size of the image formed is 1.25 cm .
Answer:
transitions in PowerPoint
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