Question Library

When the hole of a pinhole camera is diminished, the image becomes II sharp. A smaller aperture reduces the amount of light entering, leading to a sharper image, but it also requires more light for proper exposure.

Under blue light, a red rose will appear black or very dark. This is because the red pigments in the rose reflect red light and absorb other colors. Since blue light does not contain red wavelengths, no light is reflected back to the observer, causing the rose to appear dark.

When an object is placed in front of a plane mirror, the image formed is at the same distance from the mirror as the object but on the opposite side. Thus, the total distance between the object and the image is 1.5 m + 1.5 m = 3.0m

When an object is placed at the centre of curvature of a concave mirror, an image that is the same as the object will be formed. Therefore, the height of the image is 2.0 cm.

The main disadvantage of a convex mirror as a driving mirror is that it distorts the perception of distance, making them appear farther away than they are.

(a) The incident light rays must be parallel and close to the principal axis (b) Spherical aberration as a defect of curved mirrors: When a wide parallel beam is incident on a spherical mirror, it is not brought to the same focus. This makes the image distorted. (c) SEE THE DIAGRAM ABOVE (d) Accommodation is the ability of the eyes to alter its focal length to form a clear image (e) Quantitatively, (i) far point = infinity, Near point = 25 cm P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\) (ii) for near point, u = 0.25 m P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) + \(\frac{1}{v}\) similarly, for far point, u = infinity(\infty) P\(_{far point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\) P\(_{far point}\) = \(\frac{1}{\infty}\) + \(\frac{1}{v}\) = 0 + \(\frac{1}{v}\) So, change in optical power \DeltaP = P\(_{near point}\) - P\(_{far point}\) \DeltaP = 4 + \(\frac{1}{v}\) - \(\frac{1}{v}\) \DeltaP = 4 dioptres.

Refractive index = \(\frac{\text{real depth}}{\text{apparent depth}}\) Real depth = refractive index x apparent depth Apparent depth = 10cm Refractive index = 1.33 Real depth = 1.33 x 10 = 13.3cm.

When light travels from a less dense medium (like air) to a denser medium (like glass or water), its speed decreases due to the higher optical density of the new medium. The frequency of light remains constant when it passes from one medium to another. Although the speed changes, the frequency is determined by the source of the light and does not change with the medium.

Chromatic aberration occurs because different wavelengths of light are refracted by different amounts when passing through a lens. Chromatic aberration can be corrected by using a combination of diverging (concave) and converging (convex) lenses, which help focus different colors of light to the same point.

Twice the focal length of the lens: this is because, for a real image to be formed, the object must be placed outside the focal length of the lens, typically at a distance greater than twice the focal length.

Diameter = 20cm The radius of curvature, R = 10cm But, focal length, f = \(\frac{R}{2} = \frac{10}{2}\) = 5 cm. Because the inside portion is polished, light reflects off the hollow, caved-in interior. A mirror that reflects light from its inner surface is concave.

The formation of a rainbow is primarily due to dispersion, as it involves the separation of light into its constituent colors when it passes through water droplets in the atmosphere, it is refracted and dispersed due to varying wavelengths.

distance between two successive crests of a water wave = \(\lambda\) = 0.25m Frequency (f) is defined as the number of cycles or oscillations that occur in one second = four complete vertical oscillations in one second = 4Hz Speed of wave (v) = f x \(\lambda\) = 0.25 x 4 = 1m/s

For a sonometer, F \(\propto\) \(\sqrt{T}\) F = k \(\sqrt{T}\) F\(_1\) = 50Hz, F\(_2\) = ? T\(_1\) = T, T\(_2\) = 4T \(\frac{F_1}{\sqrt{T_1}} = \frac{F_2}{\sqrt{T_2}}\) \(\frac{50}{\sqrt{T}} = \frac{F_2}{\sqrt{4T}}\) F\(_2\) = \(\frac{50 \times 2 \times\sqrt{T}}{\sqrt{T}}\) F\(_2\) = 100Hz

Beat frequency = f\(_1\) - f\(_2\) f\(_1\) = 6Hz f\(_2\) = 4Hz Beat frequency = f\(_1\) - f\(_2\) = 6 - 4 = 2Hz