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

Given:

Object height (size of the candle), h = 2.5 \text{ cm}

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

Radius of curvature of the concave mirror, R = -36 \text{ cm} (concave mirror's R is negative)

The focal length of the concave mirror is half of its 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, which is 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 in front of the mirror, on the same side as the object. This means the image is real and a screen should be placed at this distance to obtain a sharp image.

Location of the image: The screen should be placed 54 cm away from the mirror.

Now, let's find the nature and size of the image using the magnification formula:

m = \frac{\text{Image height (h')}}{\text{Object height (h)}} = -\frac{v}{u}

We can find the image height (h'):

h' = m \times h = \left(-\frac{v}{u}\right) \times h

Substitute the values:

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}

Nature and size of the image: The image height is 5 cm. The negative sign for magnification (and thus for h') indicates that the image is inverted relative to the object. Since the image is formed in front of the mirror (v is negative), it is a real image. The image is magnified as |m| = |-2| = 2, meaning the image is twice the size of the object.

Effect of moving the candle closer: If the candle is moved closer to the mirror (i.e., u decreases, becoming closer to f), the image distance v will increase in magnitude and move further 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 (size of the needle), h = 4.5 \text{ cm}

Object distance, u = -12 \text{ cm} (object is placed in front of the mirror)

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

Let the image distance be v.

We use the mirror formula to find the image distance:

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

Rearrange the formula 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, which is 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 sign for v indicates that the image is formed behind the mirror. This means the image is virtual and erect.

Location of the image: The image is located approximately 6.67 cm behind the mirror.

Now, let's calculate the magnification (m):

m = -\frac{v}{u}

Substitute the values of v and u:

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

m = \frac{20}{3 \times 12} = \frac{20}{36} = \frac{5}{9}

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

Effect of moving the needle farther: As the needle (object) is moved farther from the convex mirror (i.e., u increases in magnitude, moving away from the mirror), the image distance v will decrease, and the image will move closer to the focal point (f) from the mirror. The image will remain virtual, erect, and diminished, but it will become even smaller.