5. Let A, B be the centres of two circles of equal radii: draw them so that each one
them passes through the centre of the other. Let them intersect at C and D.
Answers
Step-by-step explanation:
Given:
Height of object, \sf h_oh
o
= 5 cm
Object distance, u = - 20 cm
Radius of curvature, R = 30 cm
Focal length, f = R/2 = 30/2 = 15 cm
To find:
Position of image, it's nature and size?
Solution:
\begin{gathered}\dag\;{\underline{\frak{Using\;mirror\;formula\;:}}}\\ \\\end{gathered}
†
Usingmirrorformula:
\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}\\ \\\end{gathered}
⋆
f
1
=
v
1
+
u
1
\begin{gathered}:\implies\sf \dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{- 20}\\ \\\end{gathered}
:⟹
15
1
=
v
1
+
−20
1
\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{ - 20}\\ \\\end{gathered}
:⟹
v
1
=
15
1
−
−20
1
\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{20}\\ \\\end{gathered}
:⟹
v
1
=
15
1
+
20
1
\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{7}{60}\\ \\\end{gathered}
:⟹
v
1
=
60
7
\begin{gathered}:\implies\sf v = \dfrac{60}{7}\\ \\\end{gathered}
:⟹v=
7
60
\begin{gathered}:\implies{\boxed{\frak{\pink{v = 8.6\;cm}}}}\;\bigstar\\ \\\end{gathered}
:⟹
v=8.6cm
★
\begin{gathered}\therefore\;{\underline{\sf{Image\;distance\;is\; \bf{8.6\;cm}.}}}\\ \\\end{gathered}
∴
Imagedistanceis8.6cm.
★ Nature of image:
Image is virtual and erect.
Image formed behind the mirror.
⠀━━━━━━━━━━━━━━━━━━━━━━━━━
Size of image,
\begin{gathered}\dag\;{\underline{\frak{Using\; Formula\;of\;magnification\;:}}}\\ \\\end{gathered}
†
UsingFormulaofmagnification:
\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{h_i}{h_o} = \dfrac{v}{u}}}}}\\ \\\end{gathered}
⋆
h
o
h
i
=
u
v
\begin{gathered}:\implies\sf \dfrac{h_i}{5} = \dfrac{8.6}{20}\\ \\\end{gathered}
:⟹
5
h
i
=
20
8.6
\begin{gathered}:\implies\sf h_i = \dfrac{8.6}{20} \times 5\\ \\\end{gathered}
:⟹h
i
=
20
8.6
×5
\begin{gathered}:\implies{\boxed{\frak{\pink{h_i = 2.2\;cm}}}}\;\bigstar\\ \\\end{gathered}
:⟹
h
i
=2.2cm
★
\therefore\;{\underline{\sf{Height\;or\;size\;of\;image\;is\; \bf{2.2\;cm}.}}}∴
Heightorsizeofimageis2.2cm.