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Question 13 Marks

A $10\ cm$ long stick is kept horizontally in front of the concave mirror having focal length of $10\ cm$ in such a way that the end of the stick closest to the pole is at a distance of $20\ cm$. What will be the length of the image?

Answer

The stick is kept parallel to the Principal axis. Distance between $A$ and $P$ is $20\ cm$.Say $u_1 = 20\ cm.$
Hence, the other end of the stick is at distance, $u_2 = (u_1 + 10) = 30 cm$ from pole of the mirror.

$\begin{aligned} & \frac{1}{f}=\frac{1}{v_1}+\frac{1}{u_1} \\ & \frac{1}{v_1^{\prime}}=\frac{1}{f}-\frac{1}{u_1} \\ & =\frac{1}{-10}-\frac{1}{-20} \\ & =\frac{1}{20}-\frac{1}{10} \\ & =-\frac{1}{20} \\ & \therefore \quad v_1=-20 cm \\ & \text { Also, for other end of the stick, } \\ & \frac{1}{v_2}=\frac{1}{f}-\frac{1}{u_2} \\ & =\frac{1}{-10}-\frac{1}{-30} \\ & =\frac{1}{30}-\frac{1}{10} \\ & =-\frac{1}{15} \\ & \therefore \quad v_2=-15 cm \\ & \end{aligned}$

The length of the image formed ?s given by, $u = u_2 – u_1 = 15 – (-20) = 5\ cm.$
The length of the image is $5\ cm.$

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Question 23 Marks

Rajashree wants to get an inverted image of height 5 cm of an object kept at a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At what distance from the mirror should she place the screen? What will be the type of the image, and what is the height of the object?

Answer

Focal length = $f = -10 cm,$
Object distance = $u = -30 cm$
Height of the image = $h_2 = 7 cm$
To find: Height of the object = $h_1 = ?$
Image distance = u =?Formulae: $\frac{1}{v}+\frac{1}{h_2^u}=\frac{1}{f_v}$ $ M =\frac{h_2}{h_1}=-\frac{u}{u} $
Solution : $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$
\begin{aligned}
\frac{1}{v} & =\frac{1}{f}-\frac{1}{u} \\
\frac{1}{v} & =\frac{1}{-10}-\frac{1}{-30} \\
\frac{1}{v} & =\frac{-3+1}{30} \\
\frac{1}{v} & =\frac{-2}{30} \\
\frac{1}{v} & =\frac{1}{-15} cm \\
\therefore \quad v & =-15 cm
\end{aligned}
$
Rajashree has to place the screen 15 cm to the left of the mirror.
Magnification formula
$\begin{aligned} & M =\frac{h_2}{h_1}=-\frac{v}{u} \\ & h_1=\frac{-u h_2}{v} \\ & h_1=\frac{-(-30) \times(-5)}{-15} \\ & h_1=(-2) \times(-5) \\ & h_1=10 cm \end{aligned}$

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Question 33 Marks

An object 2 cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3 cm high. Find the image distance.

Answer

Formula : $M =\frac{h_2}{h_1}=-\frac{v}{u} .$. (As image is real.)
Solution: $\frac{h_2}{h_1}=-\frac{v}{u}$
$
\begin{aligned}
& \frac{-3}{2}=\frac{-v}{-16} \\
\therefore \quad & v=\frac{-16 \times 3}{2} \\
\therefore \quad & v=-24 cm
\end{aligned}
$

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Question 43 Marks

An object 4cm in height is placed at a distance of 36 cm from a concave mirror. The image is formed 18 cm in the front of the mirror. Find the height of the image.

Answer

Formula: $M =\frac{h_2}{h_1}=-\frac{v}{u} .$. (As image is real.)
Solution: $\frac{h_2}{h_1}=-\frac{v}{u}$
$
\begin{array}{ll}
\therefore & h_2=-\frac{h_1 v}{u} \\
\therefore & h_2=-\left(\frac{4 \times-18}{-36}\right) \\
\therefore & h_2=-2 cm
\end{array}
$

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Question 53 Marks

An object placed 20 cm in front of a convex mirror is found to have an image 15cm behind the mirror. Find the focal length of the mirror.

Answer

Formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Solution: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$
\begin{aligned}
& \therefore \quad \frac{1}{15}+\frac{1}{(-20)}=\frac{1}{f} \\
& \therefore \quad \frac{1}{15}-\frac{1}{20}=\frac{1}{f} \\
& \therefore \quad \frac{4-3}{60}=\frac{1}{f} \\
& \therefore \quad \frac{1}{60}=\frac{1}{f} \\
& \therefore \quad f=60 cm
\end{aligned}
$

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Question 63 Marks

An image is formed 5 cm behind a convex mirror of focal length 10 cm. At what distance is the object placed from the mirror?

Answer

Formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Solution: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$
\begin{array}{ll}
\therefore & \frac{1}{u}=\frac{1}{f}-\frac{1}{v} \\
\therefore & \frac{1}{u}=\frac{1}{10}-\frac{1}{5} \\
\therefore & \frac{1}{u}=\frac{1-2}{10} \\
\therefore & \frac{1}{u}=\frac{-1}{10} \\
\therefore & u=-10 cm
\end{array}
$

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Question 73 Marks

A coin is kept in front of two plane mirrors inclined to each other. If 3 images of the coin are seen then what is the angle A between the mirrors?

Answer

Formula: $n=\frac{360^{\circ}}{ A }-1$
Solution: $n=\frac{360^{\circ}}{ A }-1$
$
\begin{aligned}
n+1 & =\frac{360^{\circ}}{ A } \\
A & =\frac{360^{\circ}}{n+1} \\
& =\frac{360^{\circ}}{3+1} \\
& =\frac{360^{\circ}}{4} \\
A & =90^{\circ}
\end{aligned}
$

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Question 83 Marks

A bird is sitting in front of two plane mirrors inclined at an angle of 600 to each other. How many images does the bird see in the mirror?

Answer

Formula: $\quad n=\frac{360^{\circ}}{ A }-1$
Solution:
$
\begin{aligned}
& n=\frac{360^{\circ}}{ A }-1 \\
& =\frac{360^{\circ}}{60^{\circ}}-1 \\
& =6-1 \\
& n=5
\end{aligned}
$

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Question 93 Marks

4. Which type of mirrors are used in the following?
Periscope, floodlights, shaving mirror, kaleidoscope, street lights, headlamps of a car.

Answer

Objects	Type of Mirror
Periscope	Plane mirror
Floodlights	Concave mirror
Shaving mirror	Concave mirror
Kaleidoscope	Plane mirror
Street lights	Convex mirror
Head lamps of car	Concave mirror

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