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

Answer:

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:

Explanation:

$</p><p>Using \: mirrorformula:</p><p></p><p>\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}\\ \\\end{gathered}</p><p></p><p>$

$⟹−151=v1+−201</p><p></p><p>\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} - \dfrac{1}{ - 20}\\ \\\end{gathered}:⟹v1=−151−−201</p><p></p><p>$

$⟹v1=−151+201</p><p></p><p>\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 4 + 3}{60}\\ \\\end{gathered}:⟹v1=60−4+3</p><p></p><p>\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}\\ \\\end{gathered}:⟹v1=60−1</p><p></p><p>\begin{gathered}:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar\\ \\\end{gathered}:⟹v=−60cm★</p><p></p><p>$

$Image \: distanceis−60cm.</p><p>$

★ Nature of image:

Image is virtual and erect.

Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━

$Size of image,</p><p></p><p>\begin{gathered}\dag\;{\underline{\frak{Using\; Formula\;of\;magnification\;:}}}\\ \\\end{gathered}†UsingFormulaofmagnification:</p><p></p><p>\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{h_i}{h_o} = \dfrac{v}{u}}}}}\\ \\\end{gathered}⋆hohi=uv</p><p></p><p>\begin{gathered}:\implies\sf \dfrac{h_i}{5} = \cancel{ \dfrac{- 60}{20}}\\ \\\end{gathered}:⟹5hi=20−60</p><p></p><p>$

⟹h

i

=−3×5

\

$begin{gathered}:\implies{\boxed{\frak{\pink{h_i = - 15\;cm}}}}\;\bigstar\\ \\\end{gathered} </p><p>:⟹ </p><p>h </p><p>i</p><p> </p><p> =−15cm</p><p> </p><p> ★$

$∴Height \: or size \: of \: image \: is−15cm.</p><p>$

Answer 2

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

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.

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

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

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

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

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

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

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

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