Q4. When two plane mirrors are inclined at an angle of 90° to each other, how many images are formed? Q5. A…

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Q4. When two plane mirrors are inclined at an angle of 90° to each other, how many images are formed? Q5. A concave mirror has a radius of curvature 12 cm. At what distance from the mirror should an object be placed to give a virtual image three times the object size? Q6. The real image of an object placed 20 cm from a concave mirror is three times the object. Calculate the focal length of the mirror. a. 15.0 cm b. 30.0 cm c. 60.0 cm d. 60.0 cm

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Q4. When two plane mirrors are inclined at an angle of 90° to each other, how many images are formed?

Use the formula for the number of images formed by two inclined plane mirrors.
The number of images $N$ formed by two plane mirrors inclined at an angle $\theta$ is given by:
$N = \frac{360^\circ}{\theta} - 1$
Substitute the given angle into the formula.
Given $\theta = 90^\circ$ .
$N = \frac{360^\circ}{90^\circ} - 1$ $N = 4 - 1$ $N = 3$
Three images are formed.

The number of images formed is $\boxed{\text{3}}$ .

Q5. A concave mirror has a radius of curvature 12 cm. At what distance from the mirror should an object be placed to give a virtual image three times the object size?
Calculate the focal length of the concave mirror.
For a concave mirror, the focal length $f$ is half the radius of curvature $R$ .
$f = \frac{R}{2}$
Given $R = 12 \text{ cm}$ .
$f = \frac{12 \text{ cm}}{2}$ $f = 6 \text{ cm}$
Since it's a concave mirror, the focal length is positive by convention for the mirror formula (or negative if using Cartesian sign convention where light travels left to right, but for the mirror formula, $f$ is positive for concave and negative for convex). Let's use the convention where $f$ is positive for concave mirrors when applying the mirror formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .
Use the magnification formula to relate image distance ( $v$ ) and object distance ( $u$ ).
The magnification $m$ is given by $m = -\frac{v}{u}$ .
A virtual image means the magnification is positive. Given that the image is three times the object size, $m = +3$ .
$3 = -\frac{v}{u}$ $v = -3u$
Apply the mirror formula.
The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .
Substitute $f = 6 \text{ cm}$ and $v = -3u$ into the formula.
$\frac{1}{6} = \frac{1}{u} + \frac{1}{-3u}$ $\frac{1}{6} = \frac{1}{u} - \frac{1}{3u}$
Solve for the object distance $u$ .
Find a common denominator for the terms on the right side.
$\frac{1}{6} = \frac{3}{3u} - \frac{1}{3u}$ $\frac{1}{6} = \frac{3 - 1}{3u}$ $\frac{1}{6} = \frac{2}{3u}$
Cross-multiply to solve for $u$ .
$3u = 2 \times 6$ $3u = 12$ $u = \frac{12}{3}$ $u = 4 \text{ cm}$
The object should be placed at a distance of $\boxed{\text{4 cm}}$ from the mirror.

Q6. The real image of an object placed 20 cm from a concave mirror is three times the object. Calculate the focal length of the mirror.
a. 15.0 cm b. 30.0 cm c. 60.0 cm d. 60.0 cm
Identify the given values and determine the magnification.
Object distance $u = 20 \text{ cm}$ . (For mirrors, $u$ is usually taken as positive, and its sign in the formula depends on the convention. Let's use $u = +20 \text{ cm}$ and apply signs to $v$ and $f$ as needed.)
The image is real and three times the object, so the magnification $m = -3$ (real images are inverted, hence negative magnification).
Use the magnification formula to find the image distance $v$ .
$m = -\frac{v}{u}$ $-3 = -\frac{v}{20 \text{ cm}}$
Multiply both sides by $-1$ :
$3 = \frac{v}{20 \text{ cm}}$ $v = 3 \times 20 \text{ cm}$ $v = 60 \text{ cm}$
Since $v$ is positive, this means the real image is formed on the same side as the object, which is consistent with a real image formed by a concave mirror.
Apply the mirror formula to calculate the focal length $f$ .
The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ .
Substitute $u = 20 \text{ cm}$ and $v = 60 \text{ cm}$ .
$\frac{1}{f} = \frac{1}{20 \text{ cm}} + \frac{1}{60 \text{ cm}}$
Find a common denominator, which is 60.
$\frac{1}{f} = \frac{3}{60 \text{ cm}} + \frac{1}{60 \text{ cm}}$ $\frac{1}{f} = \frac{3 + 1}{60 \text{ cm}}$ $\frac{1}{f} = \frac{4}{60 \text{ cm}}$ $\frac{1}{f} = \frac{1}{15 \text{ cm}}$ $f = 15 \text{ cm}$
Select the correct option.
The calculated focal length is $15.0 \text{ cm}$ . This matches option (a).

The focal length of the mirror is $\boxed{\text{a. 15.0 cm}}$ .

Step-by-step solution

Use the formula for the number of images formed by two inclined plane mirrors.

Substitute the given angle into the formula.

Calculate the focal length of the concave mirror.

Use the magnification formula to relate image distance ( $v$ ) and object distance ( $u$ ).

Apply the mirror formula.

Solve for the object distance $u$ .

Identify the given values and determine the magnification.

Use the magnification formula to find the image distance $v$ .

Apply the mirror formula to calculate the focal length $f$ .

Select the correct option.

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