Question

8) Mark four points E, F, G, H such that no
three of them are collinear. Draw lines
through the points taking two at a time.
Answer the following questions.
(i) How many lines can you draw?
(ii) Name the lines concurrent at each of
these points.
(iii) How many lines can you draw if you
take the points three at a time?
Why?​

Answer 1

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.