Spherical Lenses, Lens Formula and Power Questions in English

(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.

(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.

(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) 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) In the given figure,a parallel beam of light enters the box from side $A$ and emerges from side $B$ as a diverging beam.
$A$ convex lens is a converging lens,which would cause the parallel rays to converge at a point.
$A$ rectangular glass slab would cause a lateral shift but the rays would remain parallel to each other.
$A$ prism would deviate the rays but would not cause them to diverge in this specific pattern.
$A$ concave lens is a diverging lens,which causes parallel rays of light to spread out (diverge) after passing through it. Therefore,a concave lens is placed inside the box.

(D) The power $P$ of a lens is given by $P = 1/f$, where $f$ is the focal length in meters.
For a convex lens, $f$ is positive, and for a concave lens, $f$ is negative.
For option $D$: $A$ convex lens has $f = +0.25 \, \text{m}$.
$P = 1 / 0.25 = +4 \, \text{D}$.
Thus, a convex lens with a focal length of $0.25 \, \text{m}$ has a power of $4 \, \text{D}$.

(N/A) Yes,the statement is correct.
$1$. To obtain a magnified virtual image,the object must be placed between the optical center $(O)$ and the principal focus $(F_1)$,i.e.,at a distance less than $20 \, cm$ from the lens.
$2$. To obtain a magnified real image,the object must be placed between the principal focus $(F_1)$ and the center of curvature $(2F_1)$,i.e.,at a distance greater than $20 \, cm$ but less than $40 \, cm$ from the lens.

(A) When the object is at a large distance (like a building outside),the rays of light coming from it are considered parallel to the principal axis.
According to the lens formula,for an object at infinity,the image is formed at the principal focus $(F)$.
Since the image of the nearby window pane was formed at $15\, cm$,the building (which is much further away) will form an image closer to the focal point of the lens.
Therefore,Sudha should move the screen towards the lens to obtain a sharp image of the building.
The approximate focal length of the lens is equal to the image distance of the distant object,which is $15\, cm$.

(A) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,expressed as $P = \frac{1}{f(m)}$.
This indicates that power is inversely proportional to the focal length $(P \propto \frac{1}{f})$.
$A$ lens with a shorter focal length has a higher power and thus a greater ability to converge light rays.
Comparing the two lenses,the lens with a focal length of $20\, cm$ has a higher power than the lens with a focal length of $40\, cm$.
Therefore,the lens with a focal length of $20\, cm$ will provide more convergent light.

(N/A) For a concave lens,the image formed is always virtual,erect,and diminished,regardless of the object's position.
$(a)$ When the object is at the focus $(F_1)$,the rays appear to diverge from a point between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and highly diminished.
$(b)$ When the object is between the focus $(F_1)$ and twice the focal length $(2F_1)$,the image is formed between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and diminished.
$(c)$ When the object is beyond twice the focal length $(2F_1)$,the image is formed between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and diminished.

(A) Given that the image is formed on a screen,it must be a real image.
For a real image,the magnification $m$ is negative. Given $m = -3$.
We know that $m = \frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance.
Since the image is on the other side of the lens,$v = +80 \, cm$.
Using the magnification formula: $-3 = \frac{80}{u}$.
Therefore,$u = -\frac{80}{3} \, cm \approx -26.67 \, cm$.
The negative sign indicates that the object is placed in front of the lens at a distance of $26.67 \, cm$.
The nature of the image is real and inverted,and the lens used is a convex lens.

(A) The power of a lens is defined as the reciprocal of its focal length in meters,given by the formula $P = \frac{1}{f(m)}$.
The $SI$ unit of power of a lens is Dioptre $(D)$.
For the first student,the focal length $f = 50\, cm = 0.5\, m$. Since the focal length is positive,the lens is a convex lens. Its power $P = \frac{1}{0.5} = +2\, D$.
For the second student,the focal length $f = -50\, cm = -0.5\, m$. Since the focal length is negative,the lens is a concave lens. Its power $P = \frac{1}{-0.5} = -2\, D$.

(A) $(i)$ The object distance $u = 50.0 - 12.0 = 38.0 \, cm$. The image distance $v = 88.0 - 50.0 = 38.0 \, cm$. Since $u = v$,the object is placed at $2f$. Thus,$2f = 38.0 \, cm$,which gives the focal length $f = 19.0 \, cm$.
$(ii)$ If the candle is shifted to $31.0 \, cm$,the new object distance $u' = 50.0 - 31.0 = 19.0 \, cm$. Since the object is now placed at the focus $(u' = f = 19.0 \, cm)$,the image will be formed at infinity.

(N/A) First,calculate the focal length $(f)$ of the lens:
Object distance $u = 12.0 - 50.0 = -38.0\, cm$
Image distance $v = 88.0 - 50.0 = +38.0\, cm$
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{38} - \frac{1}{-38} = \frac{1}{f} \implies \frac{2}{38} = \frac{1}{f} \implies f = 19.0\, cm$
$(i)$ New object position $= 31.0\, cm$. New $u = 31.0 - 50.0 = -19.0\, cm$. Since the object is at the focus $(u = -f)$,the image will be formed at infinity.
$(ii)$ If the candle is shifted further towards the lens (i.e.,$u < f$),the object lies between the optical center and the focus. The image formed will be virtual,erect,and magnified.
$(iii)$ The ray diagram shows the object placed between $F$ and $O$,resulting in a virtual image formed on the same side as the object.

(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.

(B) lens is a transparent optical medium bounded by at least two surfaces,of which one or both surfaces are spherical. It is an optical system designed to refract light and form images.

(A) concave lens always forms a virtual,erect,and diminished image for all positions of the object placed in front of it.
Therefore,the nature of the lens is concave.

(B) convex lens forms an erect and virtual image only when the object is placed between the optical centre $(O)$ and the principal focus $(F_1)$ of the lens. In this position,the light rays diverge after refraction,and when traced backwards,they appear to meet behind the object,forming a magnified,virtual,and erect image.

(C) When an object is placed at $2F_1$ (twice the focal length) in front of a convex lens,the light rays 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 position is at twice the focal length $(2F_1)$.

(A) convex lens is characterized by its physical shape. When you touch it,you will notice that it is thicker at the centre and thinner at the edges. This curvature causes light rays to converge,which is why it is also known as a converging lens.

(B) concave lens is defined by its physical structure where the central portion is thinner than the edges.
When you touch a concave lens,you will feel that the middle part is depressed or thinner,while the peripheral edges are thicker.

(B) The $SI$ unit of the power of a lens is the Dioptre,denoted by the symbol $D$.
One Dioptre is defined as the power of a lens whose focal length is $1 \ m$.

(N/A) The two uses of a convex lens are as follows:
$(i)$ It is used in projectors to project images onto a screen.
$(ii)$ It is used in cameras to focus light onto the image sensor or film.

(N/A) concave lens is used in spectacles to correct myopia (nearsightedness),a condition where distant objects appear blurry.

(N/A) According to the Cartesian sign convention for spherical lenses:
$1$. The optical center $(O)$ of the lens is considered as the origin $(0, 0)$.
$2$. The principal axis of the lens is taken as the $x$-axis $(X'X)$ of the coordinate system.
$3$. All distances measured in the direction of the incident light are taken as positive $(+)$.
$4$. All distances measured against the direction of the incident light are taken as negative $(-)$.
$5$. Distances measured perpendicular to and above the principal axis are taken as positive $(+)$.
$6$. Distances measured perpendicular to and below the principal axis are taken as negative $(-)$.
Therefore,the object distance $(u)$ is always negative,while the image distance $(v)$ depends on the nature of the image formed.

(B) According to the sign convention for spherical lenses,all distances are measured from the optical centre of the lens.
The optical centre is the central point of the lens through which a ray of light passes without any deviation.

(A) The power of a lens $(P)$ is defined as the reciprocal of its focal length $(f)$ in meters.
The formula is given by $P = 1 / f$.
Given,$P = 5 \ D$.
Therefore,$f = 1 / P = 1 / 5 = 0.20 \ m$.
Converting meters to centimeters,$0.20 \ m = 20 \ cm$.

(A) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters, given by the formula $P = 1/f$. A plane glass sheet can be considered as a lens with an infinite focal length $(f = \infty)$. Therefore, the power is $P = 1/\infty = 0$ diopters. Thus, the power of a plane glass sheet is zero.

(B) According to the sign convention for spherical lenses,the focal length of a concave lens is always taken as negative because the focus lies in front of the lens.
Conversely,the focal length of a convex lens is taken as positive.
Since the given focal length is $-12\, cm$,the lens must be a concave lens.

(A) The magnification $(m)$ produced by a lens is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
It is also related to the object distance $(u)$ and image distance $(v)$ by the formula:
$m = \frac{h'}{h} = \frac{v}{u}$.
Note: For a lens,the magnification is positive for virtual images and negative for real images.

(N/A) The power of a lens is defined as the ability of a lens to converge or diverge a beam of light falling on it. It is mathematically expressed as the reciprocal of its focal length in meters,given by the formula $P = 1/f$ (where $f$ is in meters). The $SI$ unit of power of a lens is Dioptre $(D)$.

(B) The power of a lens is defined as the reciprocal of its focal length in meters.
Mathematically, it is expressed as $P = \frac{1}{f}$, where $f$ is the focal length measured in meters $(m)$ and $P$ is the power measured in diopters $(D)$.

(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 $+0.5 \ m$.

(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 $1 \, m$.

(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 $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$.

(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.

(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.

(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.

(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) 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.

(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.