Spherical Lenses, Lens Formula and Power Questions in English

(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:
$P = P_1 + P_2 + P_3 + ...$

(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.

(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) When an object is placed between $F$ and $2F$ in front of a convex lens,the image is formed beyond $2F$ on the other side of the lens.
Two characteristics of the image formed are:
$1$. The image is real and inverted.
$2$. The image is magnified (larger than the object).

(N/A) convex lens forms images based on the position of the object as follows:
$(a)$ When the object is at infinity: The image is formed at the focus $F_2$,real,inverted,and highly diminished.
$(b)$ When the object is placed beyond $2F_1$: The image is formed between $F_2$ and $2F_2$,real,inverted,and diminished.
$(c)$ When the object is placed at $2F_1$: The image is formed at $2F_2$,real,inverted,and of the same size as the object.
$(d)$ When the object is placed between $F_1$ and $2F_1$: The image is formed beyond $2F_2$,real,inverted,and magnified.
$(e)$ When the object is placed at $F_1$: The image is formed at infinity,real,inverted,and highly magnified.
$(f)$ When the object is placed between the optical centre $O$ and focus $F_1$: The image is formed on the same side of the lens as the object,virtual,erect,and magnified.

(N/A) From the table of observations, it is clear that for observation $(iii)$, $u = 30 \, cm$ and $v = 30 \, cm$. This means that both $u$ and $v$ are equal to twice the focal length $(2f)$ of the convex lens.
Therefore, $2f = 30 \, cm$, which gives the focal length $f = 15 \, cm$.
$(b)$ The observation $(v)$ is incorrect. For this observation, $u = 15 \, cm$, which is equal to the focal length $(f = 15 \, cm)$. When an object is placed at the focus of a convex lens, the image is formed at infinity, not at $70 \, cm$.
$(c)$ The size of the image is equal to the size of the object when the magnification $m = 1$. Since $m = \frac{v}{u}$, the size of the image will be equal to the size of the object if $v = u$. This occurs in observation $(iii)$, where $u = 30 \, cm$ and $v = 30 \, cm$.

$(D)$ The power of a lens is defined as the reciprocal of its focal length in meters. It represents the ability of a lens to converge or diverge rays of light. Its $SI$ unit is dioptre $(D)$.
$(b)$ Given that the image is real, inverted, and of the same size as the object, the object must be placed at $2f$ (twice the focal length) from the lens, and the image is also formed at $2f$ on the other side.
Given $v = 50\, cm$, therefore $2f = 50\, cm$, which implies $f = 25\, cm = 0.25\, m$.
The needle is placed at $50\, cm$ in front of the lens.
The power of the lens $P = \frac{1}{f(\text{in meters})} = \frac{1}{0.25} = +4\, D$.

(A) Power of a lens is defined as the reciprocal of its focal length in metres. It represents the ability of a lens to converge or diverge light rays.
$SI$ unit: The $SI$ unit of power is dioptre $(D)$. One dioptre is the power of a lens whose focal length is $1\, m$.
For lens $A$: Focal length $f_A = +10\, cm = +0.1\, m$. Since the focal length is positive,it is a convex (converging) lens. Power $P_A = 1 / f_A = 1 / (+0.1\, m) = +10\, D$.
For lens $B$: Focal length $f_B = -10\, cm = -0.1\, m$. Since the focal length is negative,it is a concave (diverging) lens. Power $P_B = 1 / f_B = 1 / (-0.1\, m) = -10\, D$.
$A$ convex lens forms a virtual and magnified image when the object is placed between the optical centre and the principal focus $(f = 10\, cm)$. Since the object is placed at $8\, cm$,which is less than the focal length,lens $A$ will form a virtual and magnified image.

(N/A) Given: Focal length $f = +10 \, cm$ and object distance $u = -8 \, cm$.
According to the lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Rearranging for $v$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
Substituting the values:
$\frac{1}{v} = \frac{1}{10} + \frac{1}{-8} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = -\frac{1}{40}$
Therefore,$v = -40 \, cm$.
Hence,we conclude that:
$(a)$ The image is formed at $40 \, cm$ from the lens on the same side as the object.
$(b)$ The image is virtual and erect.
$(c)$ The image is magnified,i.e.,larger in size than the object.

(N/A) Given: Focal length $f = +20 \, cm$,Object distance $u = -30 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60}$
Therefore,$v = +60 \, cm$.
The image is formed at a distance of $60 \, cm$ on the other side of the lens. The ray diagram is as shown below.

(N/A) $(i)$ The object is placed between $F_1$ and $2F_1$ of the lens.
$(ii)$ The object is placed between the optical center $O$ and the principal focus $F_1$.
$(b)$ Refer to the provided ray diagrams.
$(c)$ $(i)$ The focal length of the lens remains unchanged because the curvature and refractive index of the lens material remain the same.
$(ii)$ The intensity of the image decreases because the aperture of the lens is reduced,allowing less light to pass through the lens.

(N/A) Given: Image distance $v = 100 \, cm$. Since the image is real and inverted, magnification $m = -2$.
Using the magnification formula $m = \frac{v}{u}$, we have $-2 = \frac{100}{u}$, which gives $u = -50 \, cm$.
Thus, the object should be placed at a distance of $50 \, cm$ in front of the lens.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we get $\frac{1}{f} = \frac{1}{100} - \frac{1}{-50} = \frac{1+2}{100} = \frac{3}{100}$.
So, $f = \frac{100}{3} \, cm = \frac{1}{3} \, m$.
The power of the lens $P = \frac{1}{f(\text{in } meters)} = \frac{1}{1/3} = 3 \, D$.
$(b)$ The laws of refraction are:
$1$. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
$2$. Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given color and for the given pair of media, i.e., $\frac{\sin i}{\sin r} = \text{constant}$.

(N/A) Definitions:
$(i)$ Optical centre: The central point of a lens through which a ray of light passes without any deviation.
$(ii)$ Centres of curvature: The centres of the two imaginary spheres of which the lens surfaces are parts.
$(iii)$ Principal axis: An imaginary line passing through the optical centre and the centres of curvature of the lens.
$(iv)$ Aperture: The effective diameter of the circular outline of a spherical lens.
$(v)$ Principal focus: $A$ point on the principal axis where light rays parallel to the principal axis converge (in convex lens) or appear to diverge (in concave lens) after refraction.
$(vi)$ Focal length: The distance between the optical centre and the principal focus of the lens.
$(b)$ Given: Focal length $f = +12 \, cm$ (for a converging lens),Image distance $v = +48 \, cm$ (real image on the other side).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{12} = \frac{1}{48} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{48} - \frac{1}{12}$
$\frac{1}{u} = \frac{1 - 4}{48} = \frac{-3}{48} = -\frac{1}{16}$
Therefore,$u = -16 \, cm$. The object should be placed at a distance of $16 \, cm$ in front of the lens.

(D) Given: Object height $h = 5\, cm$,object distance $u = -25\, cm$,focal length $f = 10\, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{-25} + \frac{1}{10} = \frac{-2 + 5}{50} = \frac{3}{50}$.
Thus,$v = \frac{50}{3}\, cm \approx 16.67\, cm$.
The positive sign indicates that the image is formed on the other side of the lens (real image).
Magnification $m = \frac{v}{u} = \frac{h'}{h}$.
$h' = h \times \frac{v}{u} = 5 \times \frac{16.67}{-25} = -3.33\, cm$.
The negative sign of $h'$ indicates that the image is inverted.
Therefore,the image is real,inverted,diminished,and formed at a distance of $16.67\, cm$ from the lens.

(B) Given: Object distance $u = -50\, cm$,Image distance $v = -10\, cm$ (since the virtual image is formed in front of the lens).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-10} - \frac{1}{-50}$.
$\frac{1}{f} = -\frac{1}{10} + \frac{1}{50} = \frac{-5 + 1}{50} = \frac{-4}{50}$.
Therefore,$f = -\frac{50}{4} = -12.5\, cm$.
The negative sign indicates that the lens is a concave lens.

(A) The power of a lens is given by the formula $P = \frac{1}{f}$,where $f$ is the focal length in meters.
Given,power $P = 2.5 \text{ D}$.
Substituting the value in the formula: $2.5 = \frac{1}{f}$.
Therefore,$f = \frac{1}{2.5} \text{ m} = 0.4 \text{ m}$.
Converting meters to centimeters: $0.4 \text{ m} = 0.4 \times 100 \text{ cm} = 40 \text{ cm}$.
Since the power is positive $(+2.5 \text{ D})$,the focal length is also positive,which indicates that the lens is a convex lens.

(A) Given: Focal length $f = +10 \ cm$,Image distance $v = +12 \ cm$ (since the image is real and formed on the wall on the other side of the lens).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{12} - \frac{1}{u} = \frac{1}{10}$.
Rearranging for $u$: $\frac{1}{u} = \frac{1}{12} - \frac{1}{10}$.
Calculating the common denominator: $\frac{1}{u} = \frac{5 - 6}{60} = \frac{-1}{60}$.
Thus,$u = -60 \ cm$.
The negative sign indicates that the object should be placed at a distance of $60 \ cm$ in front of the lens.

(B) Given that the image is real, inverted, and of the same size as the object, the object must be placed at $2f$ from the convex lens.
Since the image is formed at a distance of $v = 50\, cm$, we have $2f = 50\, cm$.
Therefore, the focal length $f = 25\, cm = 0.25\, m$.
To obtain an image of the same size, the object must be placed at $2f$, which is $50\, cm$ in front of the lens.
The power of the lens $P$ is given by the formula $P = \frac{1}{f(\text{in meters})}$.
Substituting the value, $P = \frac{1}{0.25} = 4\, D$.

(A) When lenses are placed in contact,the effective power $(P)$ of the combination is the algebraic sum of the individual powers of the lenses.
$P = P_1 + P_2$
Given $P_1 = +3.5 D$ and $P_2 = -2.5 D$.
Therefore,$P = +3.5 D + (-2.5 D) = +1.0 D$.
The focal length $(f)$ of the combination is given by the reciprocal of the total power $(f = 1/P)$.
$f = 1 / 1.0 D = 1.0 m$.
Since the power is positive,the combination behaves as a convex lens with a focal length of $1.0 m$.

(D) Given: Focal length $f = +25\, cm$,Image distance $v = +75\, cm$ (since it is on the opposite side).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{75} - \frac{1}{u} = \frac{1}{25}$.
Rearranging for $u$: $\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1 - 3}{75} = \frac{-2}{75}$.
Thus,$u = -37.5\, cm$.
The negative sign indicates the object is placed $37.5\, cm$ in front of the lens.
Since $v$ is positive and the image is formed on the opposite side,the image is real and inverted. Also,since $|v| > |u|$ $(75 > 37.5)$,the image is magnified.

(A) Given: Object height $O = 5 \, cm$,Object distance $u = -30 \, cm$,Focal length $f = +20 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30} = \frac{3-2}{60} = \frac{1}{60}$.
Therefore,$v = 60 \, cm$. The positive sign indicates the image is formed on the other side of the lens.
Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$:
$m = \frac{I}{5} = \frac{60}{-30} = -2$.
$I = 5 \times (-2) = -10 \, cm$.
$(i)$ Position: $60 \, cm$ from the lens on the opposite side.
(ii) Nature: Real and inverted (since $v$ is positive and $I$ is negative).
(iii) Size: $10 \, cm$ tall.

(D) Given: Object height $O = 10 \, cm$,object distance $u = -20 \, cm$,focal length $f = +30 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$.
Therefore,the position of the image is $v = -60 \, cm$ (the image is formed on the same side as the object).
Using the magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{10} = \frac{-60}{-20} = 3$.
Therefore,the size of the image is $I = 30 \, cm$.
Since $v$ is negative,the image is virtual and erect. Since the height $I$ is positive and greater than $O$,the image is magnified.

(A) Given: Height of object $O = 3 \, cm$,Object distance $u = -20 \, cm$,Focal length $f = +12 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{12} + \frac{1}{-20} = \frac{5 - 3}{60} = \frac{2}{60} = \frac{1}{30}$.
Therefore,$v = +30 \, cm$. The positive sign indicates the image is formed on the other side of the lens.
Using magnification $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{3} = \frac{30}{-20} = -1.5$.
$I = 3 \times (-1.5) = -4.5 \, cm$.
The image is real,inverted,and $4.5 \, cm$ in size.

(D) Given: Object height $O = 4 \ cm$,object distance $u = -27 \ cm$,focal length $f = +18 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{18} + \frac{1}{-27} = \frac{3-2}{54} = \frac{1}{54}$.
Therefore,the image position $v = +54 \ cm$.
Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{4} = \frac{54}{-27} = -2$.
Therefore,the image size $I = 4 \times (-2) = -8 \ cm$.
The positive value of $v$ indicates that the image is formed on the other side of the lens,and the negative value of $I$ indicates that the image is real and inverted.

(A) Given: Object height $O = 3.0 \text{ cm}$,focal length $f = -7.5 \text{ cm}$,image distance $v = -5.0 \text{ cm}$.
$(i)$ Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-5.0} - \frac{1}{-7.5} = -0.2 + 0.1333 = -0.0667 \text{ cm}^{-1}$.
Thus,$u = -15 \text{ cm}$. The object is placed $15 \text{ cm}$ in front of the concave lens.
(ii) Magnification $m = \frac{I}{O} = \frac{v}{u}$.
$I = \frac{v}{u} \times O = \frac{-5.0}{-15.0} \times 3.0 = 1.0 \text{ cm}$.
Since the magnification is positive $(m = +0.33)$,the image is virtual,erect,and $1.0 \text{ cm}$ in size.

(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{18} - \frac{1}{-12} = \frac{1}{18} + \frac{1}{12}$.
Taking the $LCM$ of $18$ and $12$ as $36$: $\frac{1}{f} = \frac{2+3}{36} = \frac{5}{36}$.
Therefore,$f = \frac{36}{5} = 7.2 \, cm$.
The magnification $m$ is given by $m = \frac{v}{u} = \frac{18}{-12} = -1.5$.
Since the magnitude of magnification $|m| = 1.5 > 1$,the image is magnified.

(D) $(i)$ For power $P = +2 \text{ D}$,the lens is a convex lens. Focal length $f = \frac{1}{P} = \frac{1}{+2} = +0.5 \text{ m} = +50 \text{ cm}$.
$(ii)$ For power $P = -4 \text{ D}$,the lens is a concave lens. Focal length $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$.
$(b)$ Given: Power $P = -4 \text{ D}$,object distance $u = -100 \text{ cm}$.
Focal length $f = \frac{1}{P} = -25 \text{ cm}$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-25} + \frac{1}{-100} = \frac{-4 - 1}{100} = \frac{-5}{100} = -\frac{1}{20}$.
Therefore,$v = -20 \text{ cm}$. The image is formed at a distance of $20 \text{ cm}$ on the same side as the object.

(A) Given: Object height $O = 2.0 \text{ cm}$,Focal length $f = +10 \text{ cm}$,Object distance $u = -15 \text{ cm}$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{3 - 2}{30} = \frac{1}{30}$.
Therefore,$v = +30 \text{ cm}$. The positive sign indicates the image is formed on the other side of the lens.
Magnification $m = \frac{v}{u} = \frac{I}{O}$.
$I = \frac{v}{u} \times O = \frac{30}{-15} \times 2.0 = -4 \text{ cm}$.
The image is real,inverted,and $4 \text{ cm}$ tall,formed at a distance of $30 \text{ cm}$ from the lens.

(N/A) The beam of light must be passing through the optical centre of the lens.
When a light ray passes through the optical centre of a lens,it does not undergo any refraction or deviation.
This is because the central part of the lens acts like a very thin glass slab,allowing the light to pass straight through without changing its path.

(N/A) $(i)$ $A$ convex lens can produce a magnified and virtual image when the object is placed between the optical center and the principal focus.
$(ii)$ $A$ concave lens always produces a diminished and virtual image for any position of the object.

(A) Yes,it is possible. The focal length $f$ of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\frac{n_l}{n_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
If the refractive index of the surrounding medium $(n_m)$ is greater than the refractive index of the lens material $(n_l)$,the term $(\frac{n_l}{n_m} - 1)$ becomes negative.
This causes the lens to change its nature: a convex lens (convergent) will behave as a concave lens (divergent),and vice versa.

(B) Hold the given piece of glass over some printed matter.
$(i)$ If the letters appear magnified,the given piece is a convex lens.
$(ii)$ If the letters appear diminished,the given piece is a concave lens.
$(iii)$ If the letters appear to be of the same size,then it is a plane glass plate.

(N/A) $(i)$ When $n_{1} > n_{2}$,the lens acts as a denser medium compared to the surroundings. Therefore,the concave lens behaves as a diverging lens,and the parallel rays diverge after passing through it.
$(ii)$ When $n_{1} = n_{2}$,the lens and the surrounding medium have the same refractive index. There is no change in the optical density,so the light rays pass through the lens without any deviation or refraction.
$(iii)$ When $n_{1} < n_{2}$,the lens acts as a rarer medium compared to the surroundings. In this case,the nature of the lens reverses,and the concave lens behaves as a converging lens,causing the parallel rays to converge after passing through it.

(N/A) $(i)$ The refracted ray $BD$ is shown in the figure.
$(ii)$ The point through which the refracted ray passes on the principal axis is known as the principal focus (denoted as $F_2$ in the figure) of the lens.

(N/A) When an object is placed at the principal focus $(F)$ of a convex lens,the refracted rays become parallel to each other and meet at infinity. This results in the formation of a real,inverted,and highly enlarged image at infinity.
Therefore,the object should be placed at a distance of $10\, cm$ from the optical centre of the lens.
The ray diagram is as shown below.

(N/A) $(i)$ The magnification of the image formed is unity (or $1$).
$(ii)$ The image formed is $(1)$ real and $(2)$ inverted,and it is formed at the same position as the object.
$(iii)$ The distance between the object and the optical centre of the lens is called the focal length of the lens.

(CONCAVE LENS) Since the incident ray of light parallel to the principal axis is diverged by the lens,the lens used is a concave lens.
To complete the ray diagram:
$1$. Draw a concave lens at the position indicated.
$2$. Extend the diverged ray backwards until it appears to pass through the focus $F_2$.
$3$. Draw a second ray from the top of the object $B$ passing through the optical center of the lens. This ray will pass undeviated.
$4$. The point where these two rays appear to intersect gives the position of the image $A'B'$.
The completed figure is as shown below.

(N/A) $(i)$ The lens formed by the combination is a concave lens.
$(ii)$ The lens is a diverging lens.
$(iii)$ $A$ concave lens causes parallel rays of light to diverge after refraction. The rays appear to originate from a single point on the principal axis,known as the principal focus. This is illustrated in the provided diagram.

(N/A) The object $AB$ is placed between the optical center $O$ and the principal focus $F_{1}$ of the convex lens.
$(i)$ $A$ ray of light starting from $B$ and passing through the optical center $O$ goes straight without any deviation.
$(ii)$ $A$ ray of light starting from $B$ and moving parallel to the principal axis passes through the focus $F_{2}$ after refraction by the lens.
$(iii)$ The two refracted rays appear to diverge from a point $B'$. By dropping a perpendicular from $B'$ to the principal axis,we get the point $A'$. Thus,$A'B'$ is the virtual image of the object $AB$.
$(iv)$ The image formed is virtual and erect.

(N/A) $(i)$ The distance $OS$ is called the focal length of the lens.
$(ii)$ The rays will still return to the same point $S$. When the mirror $M$ is placed in contact with the convex lens $L$,the rays from the point source $S$ (placed at the focus) pass through the lens and become parallel to the principal axis. These parallel rays strike the plane mirror $M$ normally and reflect back along the same path. After passing through the lens again,they converge at the point source $S$ itself.

(N/A) The lens used is a concave lens because it forms an erect,virtual,and diminished image for an object placed anywhere in front of it.
$(i)$ The outline of the concave lens is drawn at the position $L$.
(ii) $A$ ray from $B$ passing through the optical center $O$ goes undeviated.
(iii) $A$ ray from $B$ parallel to the principal axis,after refraction,appears to diverge from the focus $F_2$.
(iv) The final image $A'B'$ is formed between the optical center $O$ and the focus $F_2$ on the same side as the object.

(N/A) The lens is a convex lens.
Since the image is virtual,erect,and magnified,the object must be placed between the optical center and the focus of the convex lens.
The lens is situated between the object $O$ and the image $I.$
To locate the lens,draw a line from the top of the object to the top of the image and extend it to intersect the principal axis; this point is the optical center of the lens.
Draw a vertical line through this point to represent the lens.
To locate the focus $F,$ draw a ray from the top of the object parallel to the principal axis,which will pass through the focus after refraction. Extend this refracted ray backward to meet the image point. The point where the parallel ray meets the principal axis after refraction is the focus $F.$

(N/A) Power is related to the focal length as $P = \frac{1}{f}$.
An increase in power means a decrease in focal length.
Since the sunlight comes from infinity,the image is formed at the focal point of the lens.
If the power is doubled $(P' = 2P)$,the new focal length becomes half of the original focal length $(f' = \frac{f}{2})$.
Therefore,to focus the sunlight on the card,she will have to move the lens closer to the card,specifically at a distance equal to half of the previous focal length.

(N/A) convex lens forms a virtual and erect image when the object is placed between the optical centre $(O)$ and the principal focus $(F_1)$.
To explain this,consider two rays originating from the top of the object $(B)$: one ray moves parallel to the principal axis and passes through the focus $(F_2)$ after refraction,while the other ray passes through the optical centre $(O)$ without deviation.
When these two refracted rays are extended backwards,they appear to meet at a point $(B')$ on the same side of the lens as the object.
Thus,the image $(A'B')$ formed is virtual,erect,and magnified,and it lies on the same side as the object.

(C) For a convex lens,when an object is placed at the centre of curvature $(2F_1)$,the image formed is real,inverted,and of the same size as the object.
This image is formed at the centre of curvature $(2F_2)$ on the other side of the lens.
Therefore,the correct position is at the centre of curvature.

(D) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,given by the formula $P = \frac{1}{f(m)}$.
Since $P \propto \frac{1}{f}$,a lens with a smaller focal length will have a higher power.
Comparing the given focal lengths: $10 \, cm < 20 \, cm < 25 \, cm < 50 \, cm$.
Therefore,the lens with the focal length of $10 \, cm$ will have the maximum power $(P = \frac{1}{0.1} = 10 \, D)$.

(D) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,given by the formula: $P = \frac{1}{f(m)}$.
Given,$P = +5.0 \,D$.
Substituting the value in the formula: $5.0 = \frac{1}{f}$.
Therefore,$f = \frac{1}{5.0} \,m = 0.2 \,m$.
To convert the focal length into centimeters,multiply by $100$: $f = 0.2 \times 100 \,cm = +20 \,cm$.
Since the power is positive,the lens is a convex lens,and its focal length is positive.

(B) convex lens is a converging lens. When light rays traveling parallel to the principal axis strike the surface of a convex lens,they undergo refraction. After passing through the lens,these rays converge at a specific point on the principal axis. This specific point is known as the principal focus $(F)$ of the lens.

(C) For a convex lens,when an object is placed between the principal focus $(F)$ and the center of curvature $(2F)$,the image formed is real,inverted,and magnified.
This image is formed beyond $2F$ on the other side of the lens.
Therefore,the correct position of the object is between $F$ and $2F$.