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A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is 1.39 and its mean distance from the earth is . What is the diameter of the sun's image on the paper?
From the relation
The angle of polarisation for any medium is 60o, what will be critical angle for this
By using ,
also
When the angle of incidence on a material is 60°, the reflected light is completely polarized. The velocity of the refracted ray inside the material is (in ms–1)
From Brewster's law
⇒
⇒ m/sec.
For refraction at a spherical surface, a student derives the relation $ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $ where $n_1$ and $n_2$ are refractive indices of the first and second media, respectively, $u$ is the object distance, $v$ is the image distance, and $R$ is the radius of curvature. Which of the following sign conventions is consistent with this formula as given in the NCERT text?
According to NCERT, 'Applying the Cartesian sign convention, OM = –u, MI = +v, MC = +R'. This implies that distances measured in the direction of incident light are taken as positive, while those measured in the opposite direction are negative. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis. This is consistent with the standard Cartesian sign convention for optics, as also stated in Section 9.2.1 and Point 3 of the Summary in NCERT (page 247). The given formula $ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $ is derived using this sign convention.
Light from a point source in a medium with refractive index $n_1$ falls on a spherical surface with centre of curvature C and radius R, refracting into a medium with refractive index $n_2$. For small angles, the relationship between angle of incidence $i$ and angle of refraction $r$ is given by:
The NCERT text states, 'Now, by Snell’s law $n_1 \sin i = n_2 \sin r$. Or for small angles $n_1 i = n_2 r$'. This is a direct application of Snell's Law under the small angle approximation, which is crucial for deriving the spherical refraction formula. The small angle approximation allows $\sin \theta \approx \theta$ (in radians).
An object 'O' is placed on the principal axis of a spherical refracting surface. The equation for refraction at this surface is given by $ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $. If the light travels from air ($n_1=1$) to glass ($n_2=1.5$) and the object is placed 100 cm in front of a convex spherical surface with a radius of curvature of 20 cm, where is the image formed?
Given $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm (object placed in front, so opposite to incident light direction), $R = +20$ cm (convex surface, so center of curvature is in the direction of incident light). Using the formula $ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $, we have: $ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{+20} $ which simplifies to $ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{1}{40} $. So, $ \frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5 - 2}{200} = \frac{3}{200} $. Therefore, $v = \frac{1.5 \times 200}{3} = \frac{300}{3} = +100$ cm. The positive sign indicates a real image formed on the right side (in the direction of incident light). This matches Example 9.5 from the NCERT text.
Which of the following assumptions is crucial for the derivation of the formula for image formation by a single spherical surface, $ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $?
The NCERT text states, 'As before, we take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made. In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.' This small aperture/small angle approximation (also known as the paraxial approximation) is fundamental to simplifying trigonometric relations (like $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$) to derive the linear formula for image formation.
For a spherical refracting surface, the angles of incidence ($i$) and refraction ($r$) are related to the refractive indices ($n_1$, $n_2$) and the angles $\angle NOM$, $\angle NCM$, $\angle NIM$ (as defined in NCERT Figure 9.15) by which of the following expressions, assuming small angles?
From NCERT, the derivation in Section 9.5.1 states: 'Now, for $\triangle NOC$, $i$ is the exterior angle. Therefore, $i = \angle NOM + \angle NCM$'. And for 'Similarly, $r = \angle NCM – \angle NIM$'. These relations are derived using basic geometric properties of triangles and are essential steps before applying Snell's Law and small angle approximations.
When light passes from one medium to another (refraction), what remains constant?
Though not explicitly stated in the context for refraction at spherical surfaces, the context for wave optics (Equations 10.1 to 10.7) implicitly shows that wavelength and speed change upon refraction ($v_1/v_2 = \lambda_1/\lambda_2$). The frequency of light is an intrinsic property of the source and does not change when light travels from one medium to another. Wavelength and speed change because the optical density of the medium changes. Therefore, frequency remains constant.
According to the Cartesian sign convention used in optics, if the incident light travels from left to right, how are distances treated?
Point 3 of the Summary on page 247 states: 'Cartesian sign convention: Distances measured in the same direction as the incident light are positive; those measured in the opposite direction are negative. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis.' This convention is standard for deriving and applying optical formulas.
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