Textbook - Light – Reflection and Refraction Questions in English

(N/A) Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.

(B) Radius of curvature,$R = 20\, cm$.
The relationship between the radius of curvature $(R)$ and the focal length $(f)$ of a spherical mirror is given by the formula: $R = 2f$.
To find the focal length $(f)$,we rearrange the formula: $f = \frac{R}{2}$.
Substituting the given value: $f = \frac{20\, cm}{2} = 10\, cm$.
Therefore,the focal length of the given spherical mirror is $10\, cm$.

(A) concave mirror produces an erect and enlarged image when the object is placed between the pole $(P)$ and the principal focus $(F)$ of the mirror.
In this position,the light rays appear to diverge from behind the mirror,forming a virtual image that is larger than the actual object.

(B) Convex mirrors are preferred as rear-view mirrors in vehicles for the following reasons:
$1$. They always form a virtual,erect,and diminished image of the objects placed in front of them.
$2$. They have a wider field of view compared to plane mirrors,which allows the driver to see a larger area of the traffic behind the vehicle.
$3$. This helps the driver in safe driving by providing a better perspective of the vehicles approaching from the rear.

(A) The radius of curvature $(R)$ of the convex mirror is given as $32\, cm$.
The relationship between the radius of curvature $(R)$ and the focal length $(f)$ of a spherical mirror is given by the formula: $R = 2f$.
To find the focal length $(f)$,we rearrange the formula: $f = \frac{R}{2}$.
Substituting the given value: $f = \frac{32\, cm}{2} = 16\, cm$.
Therefore,the focal length of the given convex mirror is $16\, cm$.

(A) The magnification $(m)$ produced by a spherical mirror is given by the formula:
$m = -\frac{v}{u}$
Since the image is real and magnified three times,the magnification $m = -3$ (as real images are inverted).
Given,object distance $u = -10 \, cm$ (according to sign convention).
Substituting the values in the formula:
$-3 = -\frac{v}{-10}$
$-3 = \frac{v}{10}$
$v = -3 \times 10 = -30 \, cm$.
The negative sign indicates that the image is formed in front of the mirror at a distance of $30 \, cm$.

(TOWARDS THE NORMAL) The light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium, it slows down and bends towards the normal. Since water $(n \approx 1.33)$ is optically denser than air $(n \approx 1.00)$, a ray of light travelling from air into water will bend towards the normal.

(D) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$:
$n = \frac{c}{v}$
Given:
Speed of light in vacuum $(c)$ = $3 \times 10^8 \, m \, s^{-1}$
Refractive index of glass $(n)$ = $1.50$
Rearranging the formula to find the speed of light in glass $(v)$:
$v = \frac{c}{n}$
$v = \frac{3 \times 10^8 \, m \, s^{-1}}{1.50}$
$v = 2 \times 10^8 \, m \, s^{-1}$
Therefore,the speed of light in glass is $2 \times 10^8 \, m \, s^{-1}$.

(N/A) The optical density of a medium is directly related to its refractive index. $A$ medium with a higher refractive index has a higher optical density,and a medium with a lower refractive index has a lower optical density.
$1$. Highest optical density: By comparing the refractive indices in the table,Diamond has the highest refractive index of $2.42$. Therefore,Diamond has the highest optical density.
$2$. Lowest optical density: By comparing the refractive indices in the table,Air has the lowest refractive index of $1.0003$. Therefore,Air has the lowest optical density.

(WATER) The speed of light in a medium is related to its refractive index $(n_m)$ by the formula:
$n_m = \frac{\text{Speed of light in vacuum (or air)}}{\text{Speed of light in the medium}} = \frac{c}{v}$
From this,the speed of light in the medium $(v)$ is given by:
$v = \frac{c}{n_m}$
This shows that the speed of light $(v)$ is inversely proportional to the refractive index $(n_m)$: $v \propto \frac{1}{n_m}$.
Therefore,light travels fastest in the medium with the lowest refractive index.
Comparing the given refractive indices:
- Water: $1.33$
- Kerosene: $1.44$
- Turpentine oil: $1.47$
Since water has the lowest refractive index $(1.33)$ among the three,light travels fastest in water.

(N/A) The refractive index of a medium $n_m$ is related to the speed of light in that medium $v$ by the relation:
$n_m = \frac{\text{Speed of light in air}}{\text{Speed of light in the medium}} = \frac{c}{v}$
Where,$c$ is the speed of light in vacuum or air.
The refractive index of diamond is $2.42$. This suggests that the speed of light in diamond is reduced by a factor of $2.42$ compared to its speed in air.

(N/A) The power of a lens is defined as the reciprocal of its focal length in metres.
If $P$ is the power of a lens and $f$ is its focal length in metres,then the formula is:
$P = \frac{1}{f(\text{in metres})}$
The $S.I.$ unit of power of a lens is Dioptre,which is denoted by the symbol $D$.
$1$ dioptre is defined as the power of a lens whose focal length is $1$ metre.
Therefore,$1\, D = 1\, m^{-1}$.

(N/A) When an object is placed at the centre of curvature,$2F_1$,of a convex lens,its image is formed at the centre of curvature,$2F_2$,on the other side of the lens. The image formed is inverted and of the same size as the object.
It is given that the image of the needle is formed at a distance of $50\, cm$ from the convex lens. Hence,the needle is placed in front of the lens at a distance of $50\, cm$.
Object distance,$u = -50\, cm$
Image distance,$v = 50\, cm$
Focal length $= f$
According to the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{50} - \frac{1}{-50} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$
$f = 25\, cm = 0.25\, m$
Power of the lens,$P = \frac{1}{f(\text{in meters})} = \frac{1}{0.25} = +4\, D$
Hence,the power of the given lens is $+4\, D$.

(B) The focal length of a concave lens is taken as negative by sign convention.
Given,focal length $f = -2 \,m$.
The power of a lens is defined as the reciprocal of its focal length in meters,given by the formula $P = \frac{1}{f \text{ (in meters)}}$.
Substituting the value: $P = \frac{1}{-2} = -0.5 \,D$.
The negative sign indicates that the lens is concave (diverging).
Therefore,the power of the given concave lens is $-0.5 \,D$.

(C) lens is an optical device that refracts light and allows it to pass through it.
For a material to be used as a lens,it must be transparent so that light can travel through it.
Water,glass,and plastic are transparent materials and can be used to make lenses.
Clay is an opaque material,meaning light cannot pass through it.
Therefore,clay cannot be used to make a lens.

(D) concave mirror forms a virtual,erect,and magnified image only when the object is placed between the pole $(P)$ and the principal focus $(F)$ of the mirror.
In this position,the light rays appear to diverge from behind the mirror,creating a virtual image that is larger than the object.

(A) When an object is placed at twice the focal length $(2F_1)$ in front of a convex lens,its image is formed at twice the focal length $(2F_2)$ on the other side of the lens.
This image is real,inverted,and of the same size as the object.

(B) According to the sign convention for spherical mirrors and lenses:
$1$. The focal length of a concave mirror is always negative.
$2$. The focal length of a concave lens is always negative.
Since both the spherical mirror and the thin spherical lens have a focal length of $-15\,cm$ (which is negative),both must be concave in nature.

(C) convex mirror always forms a virtual and erect image of an object,regardless of its distance from the mirror.
Similarly,a plane mirror also always forms a virtual and erect image of an object,regardless of the distance.
Since both plane and convex mirrors produce erect images for all positions of the object,the mirror is likely to be either plane or convex.

(D) convex lens produces a magnified,virtual,and erect image when an object is placed within its focal length.
Magnification $(m)$ for a lens is given by the formula $m = f / (f + u)$,where $f$ is the focal length and $u$ is the object distance.
For a given object distance,a lens with a shorter focal length provides higher magnification.
Therefore,to read small letters,a convex lens with a shorter focal length $(5 \, cm)$ is preferred over one with a longer focal length $(50 \, cm)$ to obtain a larger image.

(N/A) The range of the object distance should be between $0\, cm$ and $15\, cm$ (i.e.,between the pole $P$ and the principal focus $F$).
$A$ concave mirror produces an erect image only when the object is placed between its pole $(P)$ and the principal focus $(F)$.
The nature of the image formed is virtual and erect.
The image is larger (magnified) than the object.
The ray diagram showing the image formation is provided in the image below.

(N/A) Concave mirror.
$A$ concave mirror is used in the headlights of a car. This is because when a light source (bulb) is placed at the principal focus of a concave mirror,the light rays after reflection travel as a powerful parallel beam of light,which helps in illuminating the road at a long distance.

(CONVEX) The type of mirror used is a $Convex$ mirror.
Reason:
$1$. $Convex$ mirrors always form a virtual,erect,and diminished image of the objects placed in front of them.
$2$. They have a wider field of view compared to plane or concave mirrors,which allows the driver to see a larger area of the traffic behind the vehicle.
$3$. This helps the driver in safe driving by providing a better view of the road conditions behind.

(CONCAVE MIRROR) Concave mirror.
Concave mirrors are converging mirrors. They are used in solar furnaces because they converge the parallel rays of sunlight incident on them to a single point,known as the principal focus. This concentration of light energy produces a large amount of heat at that point,which is useful for heating purposes.

(N/A) Yes,the convex lens will form a complete image of an object,even if its one-half is covered with black paper. This can be understood by the following two cases:
Case $I$: When the upper half of the lens is covered.
In this case,rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.
Case $II$: When the lower half of the lens is covered.
In this case,rays of light coming from the object are refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.
Note: Although the image is complete,its intensity (brightness) will be reduced because fewer light rays are passing through the lens to form the image.

(N/A) Given:
Object height,$h_o = 5 \,cm$
Object distance,$u = -25 \,cm$
Focal length of converging lens,$f = +10 \,cm$
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{25} = \frac{5 - 2}{50} = \frac{3}{50}$
$v = \frac{50}{3} \approx 16.67 \,cm$
The positive value of $v$ indicates that the image is real and formed on the other side of the lens at a distance of $16.67 \,cm$.
Magnification,$m = \frac{v}{u} = \frac{16.67}{-25} \approx -0.667$
Also,$m = \frac{h_i}{h_o}$
$h_i = m \times h_o = -0.667 \times 5 \approx -3.33 \,cm$
The negative sign of $h_i$ indicates that the image is inverted. Thus,the image is real,inverted,and diminished in size.

(N/A) Focal length of concave lens,$f = -15 \,cm$ (by sign convention).
Image distance,$v = -10 \,cm$ (since the image formed by a concave lens is virtual and on the same side as the object).
According to the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values:
$\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15}$
$-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}$
$-\frac{1}{u} = \frac{-2 + 3}{30} = \frac{1}{30}$
$u = -30 \,cm$
The negative sign indicates that the object is placed $30 \,cm$ in front of the lens.

(A) Focal length of convex mirror,$f = +15\, cm$.
Object distance,$u = -10\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-10}) = \frac{1}{15} + \frac{1}{10}$.
$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$.
Therefore,$v = +6\, cm$.
The positive value of $v$ indicates that the image is formed $6\, cm$ behind the mirror.
Magnification,$m = -\frac{v}{u} = -(\frac{6}{-10}) = +0.6$.
The positive value of magnification indicates that the image formed is virtual and erect.

(B) The magnification $(m)$ produced by a mirror is defined as the ratio of the height of the image $(H_i)$ to the height of the object $(H_o)$:
$m = \frac{H_i}{H_o}$
Given that the magnification produced by a plane mirror is $+1$:
$1$. The magnitude $1$ indicates that the height of the image is equal to the height of the object $(H_i = H_o)$,meaning the image is of the same size as the object.
$2$. The positive sign $(+)$ indicates that the image formed is virtual and erect.

(N/A) Object distance,$u = -20\, cm$.
Object height,$h = 5\, cm$.
Radius of curvature,$R = 30\, cm$.
Focal length,$f = R/2 = 15\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-20}) = \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.
$v = \frac{60}{7} \approx 8.57\, cm$.
The positive value of $v$ indicates that the image is formed behind the mirror.
Magnification,$m = -\frac{v}{u} = -\frac{8.57}{-20} = 0.428$.
The positive value of magnification indicates that the image is virtual and erect.
Height of the image,$h' = m \times h = 0.428 \times 5 = 2.14\, cm$.
Thus,the image is virtual,erect,and $2.14\, cm$ in size,formed $8.57\, cm$ behind the mirror.

(A) Object distance,$u = -27\, cm$.
Object height,$h = 7\, cm$.
Focal length,$f = -18\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = \frac{-3 + 2}{54} = \frac{-1}{54}$.
Thus,$v = -54\, cm$.
The screen should be placed at a distance of $54\, cm$ in front of the mirror.
Magnification,$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$.
The negative sign of magnification indicates that the image is real and inverted.
Magnification,$m = \frac{h'}{h} \implies h' = m \times h = -2 \times 7 = -14\, cm$.
The size of the image is $14\, cm$.

(A) The power of a lens is given by the formula $P = \frac{1}{f}$,where $f$ is the focal length in meters.
Given,$P = -2.0 \, D$.
Substituting the value in the formula: $-2.0 = \frac{1}{f}$.
Therefore,$f = \frac{1}{-2.0} = -0.5 \, m$.
Converting meters to centimeters: $f = -0.5 \times 100 \, cm = -50 \, cm$.
$A$ lens with a negative focal length is a concave lens.

(B) The power of a lens is given by the formula $P = \frac{1}{f}$,where $f$ is the focal length in meters.
Given power $P = +1.5\, D$.
Substituting the value,$f = \frac{1}{1.5} = \frac{10}{15} = 0.666...\, m \approx 0.67\, m$.
Converting to centimeters,$f = 0.67 \times 100 = 67\, cm$.
$A$ lens with positive power and positive focal length is a convex lens.
$A$ convex lens is a converging lens.