With the help of a ray diagram, show the formation of image of a point object due to refraction of light at a spherical surface separating two media of refractive indices $n_1$ and $n_2 (n_2 > n_1)$ respectively. Using this diagram, derive the relation.
$\frac{\text{n}_2}{\text{v}}-\frac{\text{n}_1}{\text{u}}=\frac{\text{n}_1-\text{n}_2}{\text{R}}$
Write the sign conventions used. What happens to the focal length of convex lens when it is immersed in water?
$\frac{\text{n}_2}{\text{v}}-\frac{\text{n}_1}{\text{u}}=\frac{\text{n}_1-\text{n}_2}{\text{R}}$
Write the sign conventions used. What happens to the focal length of convex lens when it is immersed in water?
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Formula for Refraction at Spherical Surface Concave Spherical Surface:Let SPS′ be a spherical refracting surface, which separates media '1' and '2'. Medium '1' is rarer and medium '2' is denser. The refractive indices of media '1' and '2' are $n_1$ and $n_2$ respectively $(n_1 < n_2)$ Let P be the pole and C the centre of curvature and PC the principal axis of spherical refracting surface.
O is a point-object on the principal axis. An incident ray OA, after refraction at A on the spherical surface bends towards the normal CAN and moves along AB. Another incident ray OP falls on the surface normally and hence passes undeviated after refraction. These two rays, when produced backward meet at point I on principal axis. Thus I is the virtual image of O. Let angle of incidence of ray OA be i and angle of refraction be r i.e.
$\angle\text{OAC} = \text{i}\text{ and } \angle\text{NAB}= \text {r}$
Let $\angle\text{AOP}=\alpha,\angle\text{AIP}=\beta\text{ and }\angle\text{ACP}=\gamma$
In triangle OAC, $\gamma=\alpha+\text{i}\text{ or}\text{ i}=\gamma-\text{a}\dots(\text{i})$
In triangle AIC, $\gamma=\beta+\text{i}\text{ or}\text{ i}=\gamma-\text{a}\dots(\text{ii})$
From Snell's law $\frac{\sin\text{i}}{\sin\text{r}}=\frac{\text{n}_2}{\text{n}_1}\dots(\text{iii})$
If point A is very near to P, then angles $\text{i},\text{r},\alpha,\beta\ \gamma$ will be very small, therefore $\sin \text{i} = \text{i} \text{ and }\sin \text{r} =\text{r}$
Substituting values of i and r from (i) and (ii) we get
$\frac{\gamma-\alpha}{\gamma-\beta}=\frac{\text{n}_2}{\text{n}_1}\text{ or }\text{n}_1(\gamma-\alpha)=\text{n}_2(\gamma-\beta)\dots(\text{iv})$
The length of perpendicular AM dropped from A on the principal axis is h i.e. AM = h. As angles $\alpha,\ \beta\ \text{and}\ \gamma$ are very small, therefore
$\tan\alpha=\alpha,\tan \beta=\beta,\gamma=\gamma$
Substituting these values in equation (iv)
$\text{n}_1(\tan\gamma-\tan\alpha)=\text{n}_2(\tan\gamma-\tan\beta)\dots(\text{v})$
As point A is very close to P, point M is coincident with P
$\tan\alpha=\frac{\text{perpendicular}}{\text{Base}}=\frac{\text{AM}}{\text{MO}}=\frac{\text{h}}{\text{PO}}$
$\tan\beta=\frac{\text{AM}}{\text{MI}}=\frac{\text{h}}{\text{PI}},$
$\tan\gamma=\frac{\text{AM}}{\text{MC}}=\frac{\text{h}}{\text{PC}}$
Substituting this value in (v), we get
$\text{n}_1\Big(\frac{\text{h}}{\text{PC}}-\frac{\text{h}}{\text{PO}}\Big)=\text{n}_2\Big(\frac{\text{h}}{\text{PC}}-\frac{\text{h}}{\text{PI}}\Big)$
$\text{or }\frac{\text{n}_1}{\text{PC}}-\frac{\text{n}_1}{\text{PC}}=\frac{\text{n}_2}{\text{PC}}-\frac{\text{n}_2}{\text{PI}}$
Let u, v and R be the distances of object O, image I and centre of curvature C from pole P. By sign convention PO, PI and PC are negative, i.e. u = -PO, v = -PI and R = -PC
Substituting these values in (vi), we get
$\frac{\text{n}_1}{(-\text{R})}-\frac{\text{n}_1}{(-\text{u})}=\frac{\text{n}_2}{(-\text{R})}-\frac{\text{n}_2}{(-\text{v})}$
$\text{or }\frac{\text{n}_2}{\text{v}}-\frac{\text{n}_1}{\text{u}}=\frac{\text{n}_2-\text{n}_1}{\text{R}}$
Sign Conventions:
$\frac{1}{\text{f}}=(\text{n}-1)\big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\big]$
When convex lens is immersed in water, refractive index n decreases and hence focal length will increase i.e., the focal length of a convex lens increases when it is immersed in water.
O is a point-object on the principal axis. An incident ray OA, after refraction at A on the spherical surface bends towards the normal CAN and moves along AB. Another incident ray OP falls on the surface normally and hence passes undeviated after refraction. These two rays, when produced backward meet at point I on principal axis. Thus I is the virtual image of O. Let angle of incidence of ray OA be i and angle of refraction be r i.e.
$\angle\text{OAC} = \text{i}\text{ and } \angle\text{NAB}= \text {r}$
Let $\angle\text{AOP}=\alpha,\angle\text{AIP}=\beta\text{ and }\angle\text{ACP}=\gamma$
In triangle OAC, $\gamma=\alpha+\text{i}\text{ or}\text{ i}=\gamma-\text{a}\dots(\text{i})$
In triangle AIC, $\gamma=\beta+\text{i}\text{ or}\text{ i}=\gamma-\text{a}\dots(\text{ii})$
From Snell's law $\frac{\sin\text{i}}{\sin\text{r}}=\frac{\text{n}_2}{\text{n}_1}\dots(\text{iii})$
If point A is very near to P, then angles $\text{i},\text{r},\alpha,\beta\ \gamma$ will be very small, therefore $\sin \text{i} = \text{i} \text{ and }\sin \text{r} =\text{r}$
Substituting values of i and r from (i) and (ii) we get
$\frac{\gamma-\alpha}{\gamma-\beta}=\frac{\text{n}_2}{\text{n}_1}\text{ or }\text{n}_1(\gamma-\alpha)=\text{n}_2(\gamma-\beta)\dots(\text{iv})$
The length of perpendicular AM dropped from A on the principal axis is h i.e. AM = h. As angles $\alpha,\ \beta\ \text{and}\ \gamma$ are very small, therefore
$\tan\alpha=\alpha,\tan \beta=\beta,\gamma=\gamma$
Substituting these values in equation (iv)
$\text{n}_1(\tan\gamma-\tan\alpha)=\text{n}_2(\tan\gamma-\tan\beta)\dots(\text{v})$
As point A is very close to P, point M is coincident with P
$\tan\alpha=\frac{\text{perpendicular}}{\text{Base}}=\frac{\text{AM}}{\text{MO}}=\frac{\text{h}}{\text{PO}}$
$\tan\beta=\frac{\text{AM}}{\text{MI}}=\frac{\text{h}}{\text{PI}},$
$\tan\gamma=\frac{\text{AM}}{\text{MC}}=\frac{\text{h}}{\text{PC}}$
Substituting this value in (v), we get
$\text{n}_1\Big(\frac{\text{h}}{\text{PC}}-\frac{\text{h}}{\text{PO}}\Big)=\text{n}_2\Big(\frac{\text{h}}{\text{PC}}-\frac{\text{h}}{\text{PI}}\Big)$
$\text{or }\frac{\text{n}_1}{\text{PC}}-\frac{\text{n}_1}{\text{PC}}=\frac{\text{n}_2}{\text{PC}}-\frac{\text{n}_2}{\text{PI}}$
Let u, v and R be the distances of object O, image I and centre of curvature C from pole P. By sign convention PO, PI and PC are negative, i.e. u = -PO, v = -PI and R = -PC
Substituting these values in (vi), we get
$\frac{\text{n}_1}{(-\text{R})}-\frac{\text{n}_1}{(-\text{u})}=\frac{\text{n}_2}{(-\text{R})}-\frac{\text{n}_2}{(-\text{v})}$
$\text{or }\frac{\text{n}_2}{\text{v}}-\frac{\text{n}_1}{\text{u}}=\frac{\text{n}_2-\text{n}_1}{\text{R}}$
Sign Conventions:
- All the distances are measured from optical centre (P) of the lens.
- Distances measured in the direction of incident ray of light are taken positive and vice-versa.
$\frac{1}{\text{f}}=(\text{n}-1)\big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\big]$
When convex lens is immersed in water, refractive index n decreases and hence focal length will increase i.e., the focal length of a convex lens increases when it is immersed in water.
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