CBSE Class 12 Physics Chapter 9: Ray Optics and Optical Instruments NCERT Solutions

Given:

Object height, h = 2.5 \text{ cm}

Object distance, u = -27 \text{ cm} (since it's in front of the mirror)

Radius of curvature, R = -36 \text{ cm} (for a concave mirror)

The focal length f is half the radius of curvature:

f = \frac{R}{2} = \frac{-36 \text{ cm}}{2} = -18 \text{ cm}

To find the image distance (v), we use the mirror formula:

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

Rearranging the formula to solve for v:

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Substitute the given values:

\frac{1}{v} = \frac{1}{-18 \text{ cm}} - \frac{1}{-27 \text{ cm}}

Find a common denominator (54):

\frac{1}{v} = \frac{-3}{54 \text{ cm}} + \frac{2}{54 \text{ cm}} = \frac{-3 + 2}{54 \text{ cm}} = \frac{-1}{54 \text{ cm}}

Therefore, the image distance is:

v = -54 \text{ cm}

The negative sign for v indicates that the image is formed on the same side as the object, which is characteristic of a real image formed by a concave mirror. Thus, the screen should be placed 54 cm from the mirror.

Now, let's find the magnification (m) and the image size (h'):

m = \frac{h'}{h} = -\frac{v}{u}

Substitute the values to find h':

h' = -\frac{v}{u} \times h = -\left(\frac{-54 \text{ cm}}{-27 \text{ cm}}\right) \times 2.5 \text{ cm}

h' = -(2) \times 2.5 \text{ cm} = -5 \text{ cm}

The magnification is m = -2. The negative sign indicates that the image is inverted. The size of the image is 5 cm. Since the image is formed at a positive distance from the mirror (in front of it) and is inverted, it is a real image.

Nature of the image: Real and inverted.

Size of the image: 5 cm.

Distance of the screen: 54 cm from the mirror.

Effect of moving the candle closer:

If the candle is moved closer to the concave mirror (e.g., from 27 cm towards the focal point at 18 cm), the object distance u decreases. According to the mirror formula, as u decreases (approaches f), the image distance v increases (moves farther away from the mirror). Therefore, the screen would have to be moved farther away from the mirror to obtain a sharp image.

Given:

Object height, h = 4.5 \text{ cm}

Object distance, u = -12 \text{ cm} (object distance is always negative)

Focal length of the convex mirror, f = +15 \text{ cm} (focal length is positive for convex mirrors)

We need to find the image distance (v) and magnification (m).

Using the mirror formula:

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

Rearranging to solve for v:

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Substitute the given values:

\frac{1}{v} = \frac{1}{15 \text{ cm}} - \frac{1}{-12 \text{ cm}}

Simplify the expression:

\frac{1}{v} = \frac{1}{15 \text{ cm}} + \frac{1}{12 \text{ cm}}

Find a common denominator (60):

\frac{1}{v} = \frac{4}{60 \text{ cm}} + \frac{5}{60 \text{ cm}} = \frac{4 + 5}{60 \text{ cm}} = \frac{9}{60 \text{ cm}}

Therefore, the image distance is:

v = \frac{60}{9} \text{ cm} = \frac{20}{3} \text{ cm} \approx 6.67 \text{ cm}

The positive value of v indicates that the image is formed behind the mirror, which is characteristic of a virtual image. The image is located 6.67 cm from the mirror.

Now, calculate the magnification (m):

m = -\frac{v}{u}

Substitute the values:

m = -\frac{\frac{20}{3} \text{ cm}}{-12 \text{ cm}}

m = -\left(-\frac{20}{3 \times 12}\right) = \frac{20}{36} = \frac{5}{9}

The magnification is m = \frac{5}{9}. Since the magnification is positive and less than 1, the image is virtual, erect, and diminished (smaller than the object).

Location of the image: 6.67 cm behind the mirror.

Magnification: \frac{5}{9} (virtual and erect).

Effect of moving the needle farther away:

As the needle (object) is moved farther away from the convex mirror, the object distance u increases (becomes more negative). According to the mirror formula, as |u| increases, the image distance v decreases (moves closer to the focal point f). The magnification m = -v/u also decreases (approaches zero). This means the virtual image formed by a convex mirror becomes smaller and moves closer to the mirror as the object moves farther away.