Mix Examples - Light – Reflection and Refraction Questions in English

(B) Given: Magnification $(m)$ = $2$,Size of the object $(O)$ = $1\, m$.
The formula for magnification is $m = \frac{I}{O}$,where $I$ is the size of the image.
Rearranging the formula to find the image size: $I = m \times O$.
Substituting the values: $I = 2 \times 1\, m = 2\, m$.
Therefore,the size of the image is $2\, m$.

(B) The focal length of a concave lens is taken as negative by sign convention.
Given, focal length $f = -25\, cm = -0.25\, m$.
The formula for the power of a lens is $P = \frac{1}{f(\text{in meters})}$.
Substituting the value, $P = \frac{1}{-0.25} = -4\, D$.
Therefore, the power of the concave lens is $-4\, D$.

(A) Given: Power of the lens,$P = +2.0 \ D$.
The relationship between power $(P)$ and focal length $(f)$ in meters is given by the formula: $P = \frac{1}{f(m)}$.
Rearranging the formula to solve for focal length: $f = \frac{1}{P}$.
Substituting the given value: $f = \frac{1}{+2.0} = +0.5 \ m$.
Therefore,the focal length of the lens is $+0.5 \ m$.

(C) When an object is placed between the pole $(P)$ and the focus $(F)$ of a concave mirror,the light rays diverge after reflection.
These rays appear to meet behind the mirror when extended backwards.
Therefore,the image formed is virtual,erect,and magnified,and it is located behind the mirror.

(B) In a concave mirror,when an object is placed at the centre of curvature $(C)$,the light rays originating from the object strike the mirror and reflect back through the centre of curvature itself. As a result,the image is formed at the same position as the object. Therefore,the object and its image coincide at the centre of curvature $(C)$.

(B) When light travels from one medium to another, its frequency remains constant because it depends on the source of the light. However, the speed of light changes as it enters a different medium. Since the relationship between speed $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by $v = f \times \lambda$, if the speed changes and the frequency remains constant, the wavelength must change. Therefore, when light passes from a rarer to a denser medium, its wavelength decreases.

(A) When a ray of light is incident normally (perpendicularly) on a reflecting surface,it travels along the normal.
Since the angle of incidence is the angle between the incident ray and the normal,and the ray itself lies on the normal,the angle of incidence is $0^{\circ}$.
According to the law of reflection,the angle of incidence is equal to the angle of reflection.
Therefore,the angle of reflection is also $0^{\circ}$.

(A) When a light ray falls normally (perpendicularly) on the surface of a glass slab,the angle of incidence $(i)$ is $0^{\circ}$.
According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$.
Since $\sin(0^{\circ}) = 0$,it follows that $\sin(r) = 0$,which means the angle of refraction $(r)$ is $0^{\circ}$.
Therefore,the ray passes through the glass slab without any deviation.

(B) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in metres, given by the formula $P = \frac{1}{f(in \, m)}$.
For the magnitude of the power to be equal to the magnitude of the focal length, we set $P = f$.
Substituting this into the formula, we get $f = \frac{1}{f}$.
This implies $f^2 = 1$, so $f = 1 \, m$ (since focal length is a physical distance).
Therefore, the magnitude of the power of a lens is equal to its focal length when the focal length is $1 \, m$.

(A) When a ray of light passes through the centre of curvature $(C)$ of a concave mirror, it strikes the mirror surface normally (perpendicularly).
Since the ray travels along the normal to the surface at the point of incidence, the angle between the incident ray and the normal is $0^\circ$.
Therefore, the angle of incidence is $0^\circ$.

(B) An air bubble inside water acts as a concave lens.
This is because the refractive index of air $(n_a \approx 1.0)$ is less than the refractive index of water $(n_w \approx 1.33)$.
When light rays pass from the denser medium (water) into the rarer medium (air bubble),they diverge.
Since a concave lens is a diverging lens,the air bubble behaves like a concave lens.

(A) Since the power is positive $(+2 \text{ D})$,the lens is a convex lens.
The focal length is calculated as $f = \frac{1}{P} = \frac{1}{2} = 0.5 \text{ m} = 50 \text{ cm}$.
$(b)$ Since the power is negative $(-4 \text{ D})$,the lens is a concave lens.
The focal length is calculated as $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$.

(N/A) Given: Refractive index of glass $(n_g)$ = $3/2$,Refractive index of water $(n_w)$ = $4/3$,Speed of light in glass $(v_g)$ = $2 \times 10^{8} \ m \ s^{-1}$.
$(i)$ Speed of light in vacuum $(c)$:
Using the formula $n_g = c / v_g$,we get $c = n_g \times v_g$.
$c = (3/2) \times (2 \times 10^{8} \ m \ s^{-1}) = 3 \times 10^{8} \ m \ s^{-1}$.
$(ii)$ Speed of light in water $(v_w)$:
Using the formula $n_w = c / v_w$,we get $v_w = c / n_w$.
$v_w = (3 \times 10^{8} \ m \ s^{-1}) / (4/3) = (3 \times 3 \times 10^{8}) / 4 = 2.25 \times 10^{8} \ m \ s^{-1}$.

(C) We prefer a convex mirror for observing the traffic behind us because its field of view is much larger than that of a plane mirror.
It forms a virtual,erect,and diminished image of the objects behind the vehicle.
Although it gives an erroneous idea about the distance and speed of the vehicles behind us,the advantage of a wider field of view makes it the preferred choice for rear-view mirrors.

(N/A) real image is formed when reflected rays actually intersect at a point.
For a plane or convex mirror,this occurs when a converging beam of light is directed towards the mirror.
The point where these rays would have met behind the mirror acts as a virtual object.
Since the reflected rays converge in front of the mirror,they form a real image at that location.

(N/A) Linear magnification $(m)$ produced by a spherical mirror is defined as the ratio of the height of the image $(h^{\prime})$ to the height of the object $(h)$.
Mathematically,it is expressed as:
$m = \frac{\text{height of the image}}{\text{height of the object}} = \frac{h^{\prime}}{h}$
Additionally,it is related to the object distance $(u)$ and image distance $(v)$ as:
$m = \frac{h^{\prime}}{h} = -\frac{v}{u}$
Here,$u$ is the distance of the object from the mirror,and $v$ is the distance of the image from the mirror.

(C) Every part of a convex lens contributes to the formation of a complete image.
When the lower part of the lens is blackened,light rays can still pass through the remaining upper part of the lens.
These rays will converge to form a complete image of the object at the same position.
However,since the total amount of light passing through the lens is reduced,the intensity (brightness) of the image will decrease.

(N/A) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a plane mirror,the focal length $f$ is considered to be infinity $(\infty)$.
Substituting $f = \infty$ into the mirror formula,we get:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{\infty}$
Since $\frac{1}{\infty} = 0$,the equation becomes:
$\frac{1}{v} + \frac{1}{u} = 0$
$\frac{1}{v} = -\frac{1}{u}$
$v = -u$
This result shows that for a plane mirror,the image distance $v$ is equal in magnitude to the object distance $u$ but opposite in sign,which is consistent with the properties of plane mirrors (virtual image formed at the same distance behind the mirror).

(N/A) No,this is not a contradiction. The lens of the eye converges the divergent beam of light (which appears to come from the virtual image) onto the retina. This process forms a real image of the virtual object on the retina,allowing us to perceive it.

(C) The focal length of a mirror depends only on its radius of curvature $(f = R/2)$ and is independent of the surrounding medium.
Therefore, the focal length of the concave mirror remains unchanged when placed in water.
However, the focal length of a lens is determined by the Lens Maker's Formula: $1/f = (n_g/n_m - 1)(1/R_1 - 1/R_2)$, where $n_g$ is the refractive index of the lens material and $n_m$ is the refractive index of the medium.
Since the refractive index of water $(n_m \approx 1.33)$ is greater than that of air $(n_m \approx 1.0)$, the term $(n_g/n_m - 1)$ decreases, causing the focal length $(f)$ of the convex lens to increase.

(A) Yes,it is possible for a lens to change its nature depending on the refractive index of the surrounding medium.
According to the Lens Maker's Formula,the focal length $f$ is given by $\frac{1}{f} = (\frac{n_l}{n_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,where $n_l$ is the refractive index of the lens and $n_m$ is the refractive index of the medium.
If the lens is placed in a medium where $n_m > n_l$,the term $(\frac{n_l}{n_m} - 1)$ becomes negative.
Consequently,the sign of the focal length $f$ reverses,causing a convex lens (convergent) to behave as a divergent lens and a concave lens (divergent) to behave as a convergent lens.

(B) Snell's law is given by the formula $n_1 \sin(i) = n_2 \sin(r)$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
When light is incident normally (perpendicularly) on the surface,the angle of incidence $i = 0^\circ$.
Since $\sin(0^\circ) = 0$,the equation becomes $n_1(0) = n_2 \sin(r)$,which implies $\sin(r) = 0$,so $r = 0^\circ$.
In this case,the law does not provide a unique relationship between the refractive indices and the angles because the ratio $\frac{\sin(i)}{\sin(r)}$ becomes $\frac{0}{0}$,which is undefined.

(N/A) When an object is placed at $2F_1$ in front of a convex lens,the image is formed at $2F_2$ on the other side of the lens.
The image formed is of the same size as the object.
The nature of the image is real and inverted.

(D) For a concave mirror,when an object is placed between the pole $(P)$ and the focus $(F)$,the light rays diverge after reflection.
When these reflected rays are extended backwards,they appear to meet behind the mirror.
This results in the formation of a virtual,erect,and magnified (larger than the object) image.
Therefore,the correct position of the object is between the pole and the focus.

(B) When an object is placed at $2F_1$ (twice the focal length) in front of a convex lens,the light rays originating from the object pass through the lens and converge at $2F_2$ on the other side. The image formed is real,inverted,and of the same size as the object. Therefore,the correct option is $(b)$.

(N/A) To distinguish between a convex and a concave lens without touching them,hold the lens close to a page of a book and observe the printed letters through it.
$1$. If the letters appear enlarged (magnified),the lens is a convex lens.
$2$. If the letters appear diminished (smaller),the lens is a concave lens.
This happens because a convex lens acts as a magnifying glass when held close to an object,whereas a concave lens always forms a virtual,erect,and diminished image.

(N/A) We can identify the three types of mirrors by observing the image formed when we look into them from a close distance:
$(i)$ If the image is of the same size and erect, it is a $Plane$ mirror.
$(ii)$ If the image is magnified (bigger size) and erect, it is a $Concave$ mirror.
$(iii)$ If the image is diminished (smaller size) and erect, it is a $Convex$ mirror.

(N/A) When a ray of light travels from one medium to another having equal refractive indices,no refraction or bending occurs. The light continues to travel in a straight line because there is no change in the optical density between the two media.
$(b)$ The cause of refraction of light is the change in the speed of light as it travels from one medium to another. When light enters a medium with a different optical density,its velocity changes,causing it to deviate from its original path.

(N/A) $(i)$ For a shaving mirror,the object (face) should be placed between the pole $(P)$ and the principal focus $(F)$ of the concave mirror to obtain an enlarged,erect,and virtual image.
$(ii)$ In torches,the source of light (bulb) should be placed at the principal focus $(F)$ of the concave mirror to produce a parallel beam of light.
$(b)$ The mirror is a combination of different types of mirrors. $A$ convex mirror is used for the upper part to produce a small head (diminished image),and a plane mirror is used for the lower part to produce an image of normal size.

(N/A) convex lens is used as a magnifying glass.
Reason: When an object is placed between the optical center $(O)$ and the principal focus $(F_1)$ of a convex lens,it forms a virtual,erect,and magnified image on the same side of the lens as the object.
Ray Diagram: The diagram shows the object $AB$ placed between $O$ and $F_1$,resulting in a virtual,magnified image $A'B'$ formed behind the object.

(A-10, B-20, C-30, D-40) For a concave mirror with focal length $f = 15 \, cm$:
$(1)$ $A$ virtual image is formed when the object is placed between the pole and the focus $(u < 15 \, cm)$. Thus,the position is $10 \, cm$.
$(2)$ $A$ diminished real image is formed when the object is placed beyond the center of curvature $(u > 30 \, cm)$. Thus,the position is $40 \, cm$.
$(3)$ An enlarged real image is formed when the object is placed between the focus and the center of curvature $(15 \, cm < u < 30 \, cm)$. Thus,the position is $20 \, cm$.
$(4)$ An image of the same size is formed when the object is placed at the center of curvature $(u = 2R = 30 \, cm)$. Thus,the position is $30 \, cm$.

(N/A) When light travels obliquely from one medium to another,the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light.
$(b)$ $A$ negative sign for the image distance $(v)$ in the case of spherical lenses indicates that the image is virtual and erect. It also signifies that the image is formed on the same side of the lens as the object.

(N/A) To locate the image of an object formed by a concave mirror,a minimum of two rays are required.
Ray diagram: $A$ concave mirror forms a virtual,erect,and magnified image of an object when it is placed between the focus $(F)$ and the pole $(P)$ of the mirror. The rays appear to diverge from behind the mirror,forming a virtual image as shown in the diagram.

(N/A) $1.$ Concave mirror
$2.$ Convex mirror
$3.$ Concave mirror
$(b)$ Differences between a real and virtual image:

Real image	Virtual image
$1.$ It is formed when light rays actually meet at a point after reflection or refraction.	$1.$ It is formed when light rays appear to meet at a point after reflection or refraction.
$2.$ It can be obtained on a screen.	$2.$ It cannot be obtained on a screen.
$3.$ It is always inverted.	$3.$ It is always erect.

(N/A) The power of a lens is defined as the measure of the degree of convergence or divergence of light rays falling on it. It is mathematically defined as the reciprocal of its focal length in meters.
i.e.,$P = 1 / f$ (where $f$ is in meters).
The $SI$ unit of power is the dioptre $(D)$.
When two or more thin lenses are placed in contact,the total power $(P)$ of the combination is the algebraic sum of the individual powers of the lenses:
$P = P_1 + P_2 + P_3 + ...$

(C) We know that the refractive index $n$ is given by $n = \frac{\sin i}{\sin r}$.
For a constant angle of incidence $i$,the refractive index $n$ is inversely proportional to $\sin r$.
Since the angle of refraction $r$ is minimum for medium $R$ $(15^{\circ})$,the refractive index $n$ is maximum for medium $R$.
We also know that the velocity of light $v$ in a medium is given by $v = \frac{c}{n}$,where $c$ is the speed of light in a vacuum.
Since $v$ is inversely proportional to $n$,the velocity of light will be minimum where the refractive index $n$ is maximum.
Therefore,the velocity of light is minimum in medium $R$.

(N/A) The absolute refractive index of a medium is defined as the ratio of the velocity of light in a vacuum to the velocity of light in that medium. If $c$ is the velocity of light in a vacuum and $v$ is the velocity in the medium,then $n = c / v$.
If $n_{a}$ and $n_{b}$ are the absolute refractive indices of media $A$ and $B$ respectively,then $n_{a} = c / v_{a}$ and $n_{b} = c / v_{b}$.
The refractive index of medium $B$ with respect to medium $A$ $(_{a}n_{b})$ is given by:
$_{a}n_{b} = \frac{\text{Velocity of light in } A}{\text{Velocity of light in } B} = \frac{v_{a}}{v_{b}} = \frac{c/n_{a}}{c/n_{b}} = \frac{n_{b}}{n_{a}}$.
Regarding the velocity of light: The velocity of light is inversely proportional to the optical density of the medium. As the optical density of a medium increases,the velocity of light in that medium decreases,and vice versa.

(A) Given: Refractive index of water $(n)$ = $4/3 = 1.33$.
Speed of light in vacuum $(c)$ = $3 \times 10^{8} \text{ m s}^{-1}$.
We know the formula for refractive index is $n = c / v$,where $v$ is the speed of light in the medium.
Rearranging the formula to find $v$: $v = c / n$.
Substituting the values: $v = (3 \times 10^{8}) / (4/3) = (3 \times 10^{8} \times 3) / 4 = 9 \times 10^{8} / 4 = 2.25 \times 10^{8} \text{ m s}^{-1}$.
Therefore,the speed of light in water is $2.25 \times 10^{8} \text{ m s}^{-1}$.

(N/A) The radius of curvature of a spherical mirror is defined as the radius of the hollow sphere of glass of which the mirror is a part.
It is denoted by the symbol $R$.
The relationship between the radius of curvature $(R)$ and the focal length $(f)$ of a spherical mirror is given by the formula $R = 2f$,which means the radius of curvature is twice the focal length.

(N/A) In this case,the light ray travels from alcohol $(n=1.36)$ to kerosene $(n=1.44)$ and then to turpentine oil $(n=1.47)$. Since the light is moving from a rarer to a denser medium at each interface,the ray of light bends towards the normal.
$(b)$ In this case,the light ray travels from turpentine oil $(n=1.47)$ to kerosene $(n=1.44)$ and then to alcohol $(n=1.36)$. Since the light is moving from a denser to a rarer medium at each interface,the ray of light bends away from the normal.

(N/A) In the provided ray diagram,$AB$ is the incident ray parallel to the principal axis.
$B$ is the point of incidence on the concave mirror.
$CB$ is the normal to the surface at point $B$,as the line joining the center of curvature $C$ to any point on the mirror is normal to the surface at that point.
According to the laws of reflection,the angle of incidence $(i)$ is equal to the angle of reflection $(r)$.
Thus,$\angle i = \angle r$,where $\angle i$ is the angle between the incident ray $AB$ and the normal $CB$,and $\angle r$ is the angle between the reflected ray $BD$ and the normal $CB$.

(A) Given: Focal length $f = +10\, cm$,Magnification $m = +2$ (for virtual image).
Using the magnification formula $m = \frac{v}{u}$,we get $v = mu = 2u$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{10} = \frac{1}{2u} - \frac{1}{u} = \frac{1-2}{2u} = -\frac{1}{2u}$.
Solving for $u$: $2u = -10$,so $u = -5\, cm$.
The object should be placed at a distance of $5\, cm$ in front of the lens.
$(b)$ If the image formed is real,then the magnification $m$ is negative. For a real image of double the size,$m = -2$. The relation between $v$ and $u$ is given by $m = \frac{v}{u}$,so $v = mu = -2u$.

(N/A) The image formed is as shown in the ray diagram. Every part of a lens contributes to the formation of an image. Therefore,if the lower half of the lens is covered,it will still form a complete,real,and inverted image at $2F_{2}$ of the same size as the object on the other side of the lens. However,the intensity (brightness) of the image will be reduced because less light passes through the lens.
$(b)$ In the second case (uncovered lens),the image will be real,inverted,formed at $2F_{2}$,and will have the same size as the object. Its intensity will be greater than the image formed in the first case because more light rays are refracted through the entire aperture of the lens. The diagram is as shown.

(N/A) $(i)$ It is a convex lens. The image is real,inverted,and magnified,which occurs when the object is placed between $F_1$ and $2F_1$.
$(ii)$ It is a concave mirror. The image is virtual,erect,and magnified,which occurs when the object is placed between the focus $(F)$ and the pole $(P)$.

(N/A) The mirror is a convex mirror.
Ray Diagram: The provided image shows the ray diagram for a convex mirror where an object $AB$ is placed in front of it,and a virtual,erect,and diminished image $A'B'$ is formed between the pole $P$ and the principal focus $F$.
Usage: Convex mirrors are generally used as rear-view mirrors in vehicles.
Reason: They are used because they always form an erect image and provide a much wider field of view,allowing the driver to see a larger area of the traffic behind the vehicle.

(A) Given: Height of object $h = +5 \, cm$,Focal length of concave lens $f = -10 \, cm$,Object distance $u = -20 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-20} = \frac{-2 - 1}{20} = \frac{-3}{20}$.
Therefore,$v = -\frac{20}{3} \, cm \approx -6.67 \, cm$. The negative sign indicates the image is formed on the same side as the object.
Nature: Since $v$ is negative,the image is virtual and erect.
Size of image $(h')$: Using magnification $m = \frac{h'}{h} = \frac{v}{u}$.
$h' = \frac{v}{u} \times h = \frac{-20/3}{-20} \times 5 = \frac{1}{3} \times 5 = +\frac{5}{3} \, cm \approx +1.67 \, cm$.

(N/A) Radius of curvature: The radius of the hollow sphere of which the reflecting surface of the spherical mirror forms a part is called the radius of curvature. It is denoted by $R$.
Focal length: The distance between the pole $(P)$ and the principal focus $(F)$ of a spherical mirror is called its focal length. It is denoted by $f$.
In the figure,$P$ is the pole,$F$ is the principal focus,and $C$ is the center of curvature. The distance $PC$ represents the radius of curvature $(R)$,and the distance $PF$ represents the focal length $(f)$.
$(b)$ Relation: For a spherical mirror of small aperture,the radius of curvature is twice the focal length. The mathematical relation is given by:
$f = \frac{R}{2}$ or $R = 2f$

(N/A) To distinguish between a plane mirror,a convex mirror,and a concave mirror,the mirror is held near the face and the image is observed.
$1$. If the image is upright,of the same size as the object,and does not change in size when the mirror is moved,the mirror is a plane mirror.
$2$. If the image is upright and magnified,and it becomes inverted when the mirror is moved away from the face,the mirror is a concave mirror.
$3$. If the image is upright and diminished,and it remains upright even when the mirror is moved away from the face,the mirror is a convex mirror.

(N/A) When an object is moved from infinity towards the pole of a concave mirror,the image shifts from the focus to infinity and then behind the mirror.
$1$. At infinity: The image is formed at the principal focus $(F)$,real,inverted,and highly diminished.
$2$. Beyond the centre of curvature $(C)$: The image is formed between $F$ and $C$,real,inverted,and diminished.
$3$. At $C$: The image is formed at $C$,real,inverted,and of the same size as the object.
$4$. Between $C$ and $F$: The image is formed beyond $C$,real,inverted,and magnified.
$5$. At $F$: The image is formed at infinity,real,inverted,and highly magnified.
$6$. Between $F$ and the pole $(P)$: The image is formed behind the mirror,virtual,erect,and magnified.

(N/A) $(i)$ The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane.
$(ii)$ The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media. This law is also known as Snell's law. Mathematically, $\frac{\sin i}{\sin r} = \text{constant} = {_1n}_2$.