Question

an object 5 cm in length is placed at a distance of 20cm in front of concave mirror of radius of curvature 30cm.Find the position of the image‚ its nature and size.​

Answer 1

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{- 4 + 3}{60}\\ \\\end{gathered}

:⟹

v

1

=

60

−4+3

\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}\\ \\\end{gathered}

:⟹

v

1

=

60

−1

\begin{gathered}:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar\\ \\\end{gathered}

:⟹

v=−60cm

★

\begin{gathered}\therefore\;{\underline{\sf{Image\;distance\;is\; \bf{- 60\;cm}.}}}\\ \\\end{gathered}

∴

Imagedistanceis−60cm.

Answer 2

$\begin{lgathered}\begin{gathered}\sf \green{\underline{ Question:-}}\\\\\end{gathered}\end{lgathered}$

5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

$\begin{lgathered}\begin{gathered}\\\\\sf \red{\underline{Given:-}}\\\\\end{gathered}\end{lgathered}$

The measures of two adjacent angles of a parallelogram are in the ratio 3:2.

$\begin{lgathered}\begin{gathered}\\\\\sf \large \red{\underline{To \: Find:-}}\\\\\end{gathered}\end{lgathered}$

Find the measure of each of the angles of the parallelogram.

$\begin{lgathered}\begin{gathered}\\\\\sf \large \red{\underline{Solution :- }}\\\\\end{gathered}\end{lgathered}$

$\text{ \sf suppose the angles be equal to 3x and 2x}$

$\boxed{ \sf \orange{ we \: have \: ardjacent \: angles \: of \: a \: parallelogram \: = 180}}$

$\begin{lgathered}\begin{gathered}\\ \sf \underline{ \green{putting \: all \: values : }}\end{gathered}\end{lgathered}$

$\begin{lgathered}\begin{gathered}\: \\ \sf \to \: 3x + 2 x = 180\: \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \:5x = 180 \\ \\ \: \sf \to \: \: \: \: \: \: \: \: \: \: \:x \: = \frac{180}{5} \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \:x \: = \cancel{ \frac{180}{5} } \\ \\ \sf \to \: \: \: \: \: \: \: \: \: \: \purple{x = 36}\\\\\end{gathered}\end{lgathered}$

$\begin{lgathered}\begin{gathered}\sf \to \: 3x \\ \sf \to \: 3 \times 36 \\ \sf \to \red{108 }\\ \\ \\ \sf \to \: 2x \\ \sf \to \: 2 \times 36 \\ \sf \to \orange{72} \\\end{gathered}\end{lgathered}$

$\sf \large\underline{ \blue{verification }} \huge \dag$

$\begin{lgathered}\begin{gathered}\\ \\ \sf \to 3x + 2x = 180 \\ \\ \sf \to \: 3 \times 36 +2 \times 36 = 180 \\ \\ \sf \to \: 108 + 72 = 180 \\ \\ \sf \to \:180 = 180 \\ \\ \large \underline{ \pink{ \sf \: hence \: verified}} \huge \dag\end{gathered}\end{lgathered}$