7 Most VSAQ’s of Ray Optics and Optical Instruments Chapter in Inter 2nd Year Physics (TS/AP)

Table of Contents

2 Marks

VSAQ-1 : Define ‘power’ of a convex lens. What is its unit?

The unit of power of a convex lens is the diopter (D). The formula to calculate power is: Power (P) = 1/Focal Length (F), where power is measured in diopters (D) and focal length is measured in meters (m).

VSAQ-2 : A small angled prism of 4°deviates a ray through 2.48°. Find the refractive index of the prism.

To find the refractive index (n) of the prism, we can use the formula:

n = tan((A + δ)/2) / tan(A/2)

Where:

A is the angle of the prism (4° in this case).
δ is the deviation caused by the prism (2.48°).

Plugging in the values:

n = tan((4° + 2.48°)/2) / tan(4°/2) n = tan(3.24°/2) / tan(2°)

Now, calculate the values:

n ≈ tan(1.62°) / tan(2°)

Using a calculator:

n ≈ 0.0284 / 0.0349

n ≈ 0.815

So, the refractive index of the prism is approximately 0.815.

VSAQ-3 : What is ‘dispersion’? Which colour gets relatively more dispersed?

Dispersion is the phenomenon in which white light is separated into its constituent colors (spectrum) when it passes through a prism or a refracting medium. This occurs because different colors of light have varying wavelengths, leading to distinct degrees of bending or refraction.

Among the colors in the spectrum, violet gets relatively more dispersed compared to other colors. Violet light undergoes a greater deviation from its original path when passing through a prism, resulting in a larger angle of deviation for violet light compared to the other colors in the spectrum.

VSAQ-4 : What is myopia? How can it be corrected?

Myopia, also known as nearsightedness, is an eye condition where distant objects appear blurry because the eye focuses incoming light in front of the retina. To correct myopia, concave lenses are used to diverge incoming light, allowing it to focus properly on the retina. This helps individuals with myopia see distant objects clearly. While there’s no cure, corrective lenses or refractive surgery effectively manage the condition.

VSAQ-5 : What is hypermetropia? How can it be corrected?

Hypermetropia, or farsightedness, is a vision condition where distant objects are seen more clearly than nearby ones. It happens when the eyeball is too short or the cornea has too little curvature, causing the image to focus behind the retina, not directly on it, resulting in blurriness for close objects.

To correct hypermetropia, converging lenses (convex lenses) are used. These lenses help converge incoming light rays before they enter the eye, allowing the image to focus correctly on the retina. This correction enables individuals with hypermetropia to see nearby objects more clearly. Corrective lenses, such as glasses or contact lenses, are commonly prescribed to manage hypermetropia and provide clear vision for both near and distant objects.

VSAQ-6 : A concave mirror produces an image of a long vertical pin, placed 40 cm from the mirror at the position of the object. Find the focal length of the mirror.

To find the focal length (f) of a concave mirror when an object is placed 40 cm from the mirror, you can use the mirror equation:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Given that the object distance (d_o) is 40 cm, and assuming the image distance (d_i) is positive (since it will be on the same side as the object), you can rearrange the equation to solve for f:

$$f = \frac{1}{\frac{1}{d_i} – \frac{1}{40 \, \text{cm}}}$$

You’ll need to know the actual value of d_i to calculate the focal length (f) of the concave mirror.

VSAQ-7 : A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall?

To position an object so that its real image is formed on a wall by a concave mirror with a focal length of 10 cm and placed at a distance of 35 cm from the wall, the object should be placed approximately 14 cm from the mirror.