Combination of Lenses Questions in English

(B) The equivalent focal length $F$ of two thin lenses separated by a distance $d$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Here,$f_1$ is the focal length of the convex lens (positive) and $f_2$ is the focal length of the concave lens (negative).
When the lenses are put in contact,the separation distance $d = 0$.
Substituting $d = 0$ into the formula,we get:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Since $f_1 > 0$ and $f_2 < 0$,let $f_1 = f$ and $f_2 = -f'$ (where $f, f' > 0$).
$\frac{1}{F} = \frac{1}{f} - \frac{1}{f'}$
As the lenses are brought from a separated state to contact,the term $-\frac{d}{f_1 f_2}$ (which was negative since $f_1 f_2 < 0$) is removed. Effectively,the power of the combination changes such that the focal length increases compared to the separated state.

(A) The equivalent focal length $(F)$ of two thin lenses separated by a distance $d$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Given $f_1 = 0.2 \,m$,$f_2 = 0.2 \,m$,and $d = 0.5 \,m$.
Substituting these values into the formula:
$\frac{1}{F} = \frac{1}{0.2} + \frac{1}{0.2} - \frac{0.5}{(0.2)(0.2)}$
$\frac{1}{F} = 5 + 5 - \frac{0.5}{0.04}$
$\frac{1}{F} = 10 - 12.5$
$\frac{1}{F} = -2.5$
Therefore,$F = -\frac{1}{2.5} = -0.4 \,m$.

(D) When two thin lenses are placed in contact,the effective focal length $F$ is given by the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
In all three cases shown,we have two plano-convex lenses,each with focal length $f$. When they are placed in contact,the combination acts as a single lens system.
Case $(i)$: The two lenses are in contact with their curved surfaces facing the same direction. The effective focal length is $\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,so $F_1 = \frac{f}{2}$.
Case $(ii)$: The two lenses are in contact forming a biconvex lens. The effective focal length is $\frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,so $F_2 = \frac{f}{2}$.
Case $(iii)$: The two lenses are in contact with their curved surfaces facing each other. The effective focal length is $\frac{1}{F_3} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$,so $F_3 = \frac{f}{2}$.
Thus,the ratio of their effective focal lengths is $F_1 : F_2 : F_3 = \frac{f}{2} : \frac{f}{2} : \frac{f}{2} = 1 : 1 : 1$.

(C) Given,the sum of powers of two thin lenses is $P_1 + P_2 = 9 \text{ D}$.
When separated by a distance $d = 20 \text{ cm} = 0.2 \text{ m}$,the equivalent power $P$ is given by the formula:
$P = P_1 + P_2 - d P_1 P_2$
Substituting the given values:
$\frac{27}{5} = 9 - 0.2 \times P_1 P_2$
$5.4 = 9 - 0.2 \times P_1 P_2$
$0.2 \times P_1 P_2 = 9 - 5.4 = 3.6$
$P_1 P_2 = \frac{3.6}{0.2} = 18$
We have $P_1 + P_2 = 9$ and $P_1 P_2 = 18$. The quadratic equation $x^2 - (P_1+P_2)x + P_1 P_2 = 0$ becomes $x^2 - 9x + 18 = 0$.
Solving this,$(x-3)(x-6) = 0$,so $x = 3$ or $x = 6$.
Thus,the individual powers are $3 \text{ D}$ and $6 \text{ D}$.

(B) Given: Focal length of convex lens $f_{1} = 30 \,cm$, focal length of concave lens $f_{2} = -20 \,cm$, and initial image size $h_{o} = 2 \,cm$.
Since the object is at infinity, the image formed by the convex lens is at its focus, $v_{1} = 30 \,cm$.
The concave lens is placed at a distance of $26 \,cm$ from the convex lens. Thus, the image formed by the convex lens acts as a virtual object for the concave lens.
The distance of this virtual object from the concave lens is $u_{2} = v_{1} - 26 = 30 - 26 = 4 \,cm$.
Using the lens formula for the concave lens: $\frac{1}{v_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$.
Substituting the values: $\frac{1}{v_{2}} - \frac{1}{4} = \frac{1}{-20}$.
$\frac{1}{v_{2}} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$.
So, $v_{2} = 5 \,cm$.
The magnification $m$ produced by the concave lens is $m = \frac{v_{2}}{u_{2}} = \frac{5}{4} = 1.25$.
The new size of the image $h_{i} = m \times h_{o} = 1.25 \times 2 \,cm = 2.5 \,cm$.

(C) The power of the combination of lenses in contact is given by $P = P_1 + P_2$.
Given $P_1 = -1.75 D$ and $P_2 = +2.25 D$.
Therefore,$P = -1.75 + 2.25 = 0.5 D$.
The focal length $f$ is related to power $P$ by the formula $f = \frac{1}{P}$ (in meters).
$f = \frac{1}{0.5} = 2 \,m$.
Since $1 \,m = 100 \,cm$,$f = 2 \times 100 = 200 \,cm$.

(D) Let the focal lengths of the two lenses be $f_1$ and $f_2$.
Given that the lenses are in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given $F = 4 \ cm$,so $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{4}$.
This simplifies to $\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{4}$.
We are given $f_1 + f_2 = 18 \ cm$.
Substituting this into the equation: $\frac{18}{f_1 f_2} = \frac{1}{4}$,which gives $f_1 f_2 = 72 \ cm^2$.
We have the sum $f_1 + f_2 = 18$ and the product $f_1 f_2 = 72$.
These are the roots of the quadratic equation $x^2 - 18x + 72 = 0$.
Factoring the equation: $(x - 6)(x - 12) = 0$.
Thus,the focal lengths are $6 \ cm$ and $12 \ cm$.
Power $P$ is inversely proportional to focal length $(P = \frac{1}{f})$.
The lens with the lower power will have the larger focal length.
Therefore,the focal length of the lens with lower power is $12 \ cm$.

(C) For the concave lens:
$u = -30 \ cm$,$f = -20 \ cm$
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-20}$
$\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = \frac{-3-2}{60} = -\frac{5}{60} = -\frac{1}{12}$
$v = -12 \ cm$ (The image is formed $12 \ cm$ to the left of the concave lens).
For the convex lens:
The image formed by the concave lens acts as an object for the convex lens.
The distance between the lenses is $5 \ cm$.
$u = -(12 + 5) = -17 \ cm$,$f = +10 \ cm$
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{-17} = \frac{1}{10}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{17} = \frac{17 - 10}{170} = \frac{7}{170}$
$v = \frac{170}{7} \approx 24.29 \ cm$ to the right of the convex lens.
Since the convex lens is $5 \ cm$ to the right of the concave lens,the final image position relative to the concave lens is $24.29 + 5 = 29.29 \ cm$ to the right of the concave lens.

(C) For a parallel beam of light to emerge parallel from a system of two lenses, the second focal point of the first lens must coincide with the first focal point of the second lens.
Let $L_1$ be the diverging lens with focal length $f_1 = -15 \,cm$ and $L_2$ be the convergent lens with focal length $f_2$.
The parallel beam incident on $L_1$ appears to diverge from its focal point $F_1$, which is $15 \,cm$ to the left of $L_1$.
For the beam to emerge parallel from $L_2$, the light rays incident on $L_2$ must appear to come from its focal point $F_2$, which is located to the left of $L_2$ at a distance equal to its focal length $f_2$.
From the geometry of the system, the distance between the two lenses is $d = 40 \,cm$.
The focal point $F_1$ is $15 \,cm$ to the left of $L_1$. Thus, the distance of $F_1$ from $L_2$ is $15 \,cm + 40 \,cm = 55 \,cm$.
Since $F_1$ and $F_2$ must coincide for the rays to emerge parallel, the focal length of the convergent lens $f_2$ must be $55 \,cm$.

(A) For a plano-convex lens with refractive index $\mu_1$ and radius of curvature $R$,the focal length $f_1$ is given by the lens maker's formula: $1/f_1 = (\mu_1 - 1)(1/R - 1/\infty) = (\mu_1 - 1)/R$.
For a plano-concave lens with refractive index $\mu_2$ and radius of curvature $R$,the focal length $f_2$ is given by: $1/f_2 = (\mu_2 - 1)(-1/\infty - 1/R) = -(\mu_2 - 1)/R$.
When the two lenses are combined,the effective focal length $F$ is given by $1/F = 1/f_1 + 1/f_2$.
Substituting the values: $1/F = (\mu_1 - 1)/R - (\mu_2 - 1)/R = (\mu_1 - 1 - \mu_2 + 1)/R = (\mu_1 - \mu_2)/R$.
Therefore,the focal length of the combination is $F = R/(\mu_1 - \mu_2)$.

(D) Given:
Dispersive power of the first lens,$\omega_1 = 0.45$.
Dispersive power of the second lens,$\omega_2 = 0.21$.
Focal length of the second lens (convex),$f_2 = 84 \,cm$.
For an achromatic combination of two thin lenses in contact,the condition is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
$\Rightarrow \frac{\omega_1}{f_1} = -\frac{\omega_2}{f_2}$
$\Rightarrow \frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$
Substituting the given values:
$\frac{f_1}{84} = -\frac{0.45}{0.21}$
$f_1 = -\frac{45}{21} \times 84$
$f_1 = -\frac{45}{1} \times 4$
$f_1 = -180 \,cm$
Thus,the focal length of the required lens is $-180 \,cm$.

(B) Given,the power of the lenses are $P_1 = -1.6 D$ and $P_2 = +2.1 D$.
When two thin lenses are placed in contact,the total power $P$ of the combination is given by the sum of individual powers:
$P = P_1 + P_2 = -1.6 D + 2.1 D = 0.5 D$.
The relationship between the focal length $f$ (in meters) and the power $P$ (in Diopters) is $P = \frac{1}{f(m)}$.
To find the focal length in centimeters,we use $f(cm) = \frac{100}{P(D)}$.
Substituting the value of $P$:
$f = \frac{100}{0.5} = 200 cm$.
Therefore,the focal length of the combination is $200 cm$.

(A) For a glass slab,the equivalent focal length $F$ is $\infty$.
For two thin lenses of focal lengths $f_1$ and $f_2$ separated by a distance $d$,the equivalent focal length $F$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Since the combination acts as a glass slab,we set $F = \infty$,which implies $\frac{1}{F} = 0$.
Substituting this into the formula:
$0 = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
$0 = \frac{f_2 + f_1}{f_1 f_2} - \frac{d}{f_1 f_2}$
$\frac{d}{f_1 f_2} = \frac{f_1 + f_2}{f_1 f_2}$
$d = f_1 + f_2$
Thus,the distance of separation between the lenses must be $f_1 + f_2$.

(C) Given: Focal lengths of the two convex lenses are $f_1 = 20 \,cm$ and $f_2 = 30 \,cm$.
When two thin lenses are placed in contact,the equivalent focal length $F$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting the given values:
$\frac{1}{F} = \frac{1}{20} + \frac{1}{30}$
$\frac{1}{F} = \frac{3 + 2}{60} = \frac{5}{60}$
$\frac{1}{F} = \frac{1}{12}$
Therefore,$F = 12 \,cm$.

(C) For the convex lens: Object distance $u_1 = -8 \,cm$, focal length $f_1 = +4 \,cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we get $\frac{1}{v_1} = \frac{1}{4} + \frac{1}{-8} = \frac{1}{8}$, so $v_1 = +8 \,cm$. This image acts as a virtual object for the concave lens.
For the concave lens: The distance between the lenses is $d = 6 \,cm$. The image formed by the convex lens is $8 \,cm$ behind it, which is $8 - 6 = 2 \,cm$ behind the concave lens. Thus, for the concave lens, $u_2 = +2 \,cm$ and $f_2 = -4 \,cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$, we get $\frac{1}{v_2} = \frac{1}{-4} + \frac{1}{2} = \frac{1}{4}$, so $v_2 = +4 \,cm$. This means the final image is formed $4 \,cm$ to the right of the concave lens.
The object is $8 \,cm$ to the left of the convex lens. The distance between the object and the convex lens is $8 \,cm$, the distance between the lenses is $6 \,cm$, and the distance between the concave lens and the final image is $4 \,cm$. The total distance between the object and the final image is $8 + 6 + 4 = 18 \,cm$.

(B) Given: Focal length of lens $A$, $f_1 = 2 \,cm$. Focal length of lens $B$, $f_2 = 5 \,cm$. Distance between lenses $d = 14 \,cm$. Object distance for lens $A$, $u_1 = -3 \,cm$.
For lens $A$, using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$\frac{1}{v_1} - \frac{1}{-3} = \frac{1}{2}$
$\frac{1}{v_1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
$v_1 = 6 \,cm$.
The image formed by lens $A$ acts as an object for lens $B$. The distance of this image from lens $B$ is $u_2 = -(d - v_1) = -(14 - 6) = -8 \,cm$.
For lens $B$, using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$\frac{1}{v_2} - \frac{1}{-8} = \frac{1}{5}$
$\frac{1}{v_2} = \frac{1}{5} - \frac{1}{8} = \frac{8 - 5}{40} = \frac{3}{40}$
$v_2 = \frac{40}{3} \,cm$.
The distance of the final image from lens $A$ is the distance between the lenses plus the distance of the image from lens $B$:
Distance $= d + v_2 = 14 + \frac{40}{3} = \frac{42 + 40}{3} = \frac{82}{3} \,cm$.

(B) The combined focal length $F$ of three thin lenses in contact is given by the formula: $\frac{1}{F} = \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3}$.
Given $F_1 = F_2 = F_3 = 3 \,cm$,we have $\frac{1}{F} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \,cm^{-1}$.
Thus,the effective focal length is $F = 1 \,cm$.
The magnification $M$ of a simple magnifier (or a lens combination acting as one) in normal adjustment is given by $M = 1 + \frac{D}{F}$,where $D$ is the least distance of distinct vision.
Given $D = 25 \,cm$ and $F = 1 \,cm$,we get $M = 1 + \frac{25}{1} = 26$.

(A) For an achromatic combination of two thin lenses,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which implies $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.
Given the ratio of dispersive powers $\frac{\omega_1}{\omega_2} = \frac{4}{3}$,we have $\frac{f_1}{f_2} = -\frac{4}{3}$,so $f_1 = -\frac{4}{3}f_2$.
The effective focal length $F$ of the combination is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting $F = 60 \ cm$ and $f_1 = -\frac{4}{3}f_2$:
$\frac{1}{60} = \frac{1}{-(4/3)f_2} + \frac{1}{f_2} = \frac{1}{f_2} (1 - \frac{3}{4}) = \frac{1}{f_2} (\frac{1}{4})$.
Thus,$f_2 = 60 \times \frac{1}{4} = 15 \ cm$.
Then,$f_1 = -\frac{4}{3} \times 15 = -20 \ cm$.
Therefore,the focal lengths are $-20 \ cm$ and $15 \ cm$.

(D) Given:
Dispersive power of the first lens, $\omega_1 = 0.45$.
Dispersive power of the second lens, $\omega_2 = 0.21$.
Focal length of the second lens, $f_2 = 84 \,cm$.
For an achromatic combination of two thin lenses in contact, the condition is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
$\frac{\omega_1}{f_1} = -\frac{\omega_2}{f_2}$
$\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$
Substituting the values:
$f_1 = -f_2 \times \left( \frac{\omega_1}{\omega_2} \right)$
$f_1 = -84 \times \left( \frac{0.45}{0.21} \right)$
$f_1 = -84 \times \left( \frac{45}{21} \right)$
$f_1 = -84 \times \frac{15}{7}$
$f_1 = -12 \times 15 = -180 \,cm$.
Thus, the focal length of the lens is $-180 \,cm$.

(D) An achromatic doublet is a combination of two lenses made of different materials designed to minimize chromatic aberration.
For an achromatic doublet,the condition for the combination to be achromatic is given by the equation:
$\omega_1 P_1 + \omega_2 P_2 = 0$
where $P_1$ and $P_2$ are the powers of the two lenses and $\omega_1$ and $\omega_2$ are their respective dispersive powers.
This equation implies that the sum of the product of their powers and dispersive powers must be equal to zero.

(C) The image of an object in white light formed by a single lens is usually coloured and blurred because the focal length of the lens varies with the wavelength of light. This defect is known as chromatic aberration.
An achromatic combination of lenses is designed to minimize or eliminate this chromatic aberration.
By combining two lenses of different materials (typically a convex lens of crown glass and a concave lens of flint glass),the dispersion caused by one lens is cancelled by the other.
Therefore,the resulting image is free from chromatic aberration,meaning the image quality is not affected by the variation of the refractive index with the wavelength of light.

(B) From the diagram,the object $O$ is at $30 \ cm$ from lens $A$. Lens $B$ is at $5 \ cm$ from lens $A$,and lens $C$ is at $10 \ cm$ from lens $B$.
For lens $A$: $u_A = -30 \ cm$,$f_A = +10 \ cm$.
Using the lens formula $\frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f_A}$:
$\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{10} \Rightarrow \frac{1}{v_A} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \Rightarrow v_A = 15 \ cm$.
The image formed by lens $A$ acts as an object for lens $B$. The distance of this image from lens $B$ is $u_B = +(v_A - 5) = +(15 - 5) = +10 \ cm$.
For lens $B$: $u_B = +10 \ cm$,$f_B = -10 \ cm$.
Using the lens formula: $\frac{1}{v_B} - \frac{1}{10} = \frac{1}{-10} \Rightarrow \frac{1}{v_B} = 0 \Rightarrow v_B = \infty$.
The rays emerging from lens $B$ are parallel,so they act as an object at infinity for lens $C$.
For lens $C$: $u_C = \infty$,$f_C = +30 \ cm$.
Using the lens formula: $\frac{1}{v_C} - \frac{1}{\infty} = \frac{1}{30} \Rightarrow v_C = 30 \ cm$.
The final image is formed at $30 \ cm$ from lens $C$. The total distance of lens $C$ from the object is $30 + 5 + 10 = 45 \ cm$. The final image is at $30 \ cm$ from lens $C$,so the distance from the object is $45 + 30 = 75 \ cm$.

(D) For the first lens:
$f_1 = +10 \ cm, u_1 = -30 \ cm$.
Applying the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.
So,$v_1 = +15 \ cm$.
This image acts as an object for the second lens. The distance between the first and second lens is $5 \ cm$,so the object distance for the second lens is $u_2 = +(15 - 5) = +10 \ cm$.
For the second lens $(f_2 = -10 \ cm)$:
$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-10} + \frac{1}{10} = 0$.
So,$v_2 = \infty$.
The rays emerging from the second lens are parallel to the principal axis.
These parallel rays fall on the third lens $(f_3 = +30 \ cm)$. Parallel rays incident on a convex lens converge at its focus.
Therefore,the final image is formed at a distance of $30 \ cm$ from the third lens.

(B) For the first lens:
Using the lens formula $\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$,where $f_1 = 2 \ m$ and $u_1 = -4 \ m$:
$\frac{1}{2} = \frac{1}{v_1} - \frac{1}{-4} \Rightarrow \frac{1}{v_1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \Rightarrow v_1 = 4 \ m$.
This image is formed $4 \ m$ to the right of the first lens.
For the second lens:
The distance between the lenses is $3 \ m$. The image formed by the first lens acts as an object for the second lens.
Since the image is $4 \ m$ to the right of the first lens and the second lens is $3 \ m$ to the right of the first,the image is $1 \ m$ to the right of the second lens.
Thus,the object distance for the second lens is $u_2 = +1 \ m$ (virtual object).
Using the lens formula $\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}$,where $f_2 = 1 \ m$ and $u_2 = +1 \ m$:
$\frac{1}{1} = \frac{1}{v_2} - \frac{1}{1} \Rightarrow \frac{1}{v_2} = 1 + 1 = 2 \Rightarrow v_2 = 0.5 \ m$.
The final image is $0.5 \ m$ to the right of the second lens.
The total distance from the source is $4 \ m$ (to first lens) $+ 3 \ m$ (between lenses) $+ 0.5 \ m$ (from second lens) $= 7.5 \ m$.

(D) For the first condition,using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
Given $f_1 = 10 \ cm$ and $u = -30 \ cm$.
$\frac{1}{10} = \frac{1}{v} - \frac{1}{-30} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.
So,$v = 15 \ cm$.
For the second condition,when the concave lens is placed in contact,the new image distance $v' = v + 45 \ cm = 15 + 45 = 60 \ cm$.
The object distance $u$ remains $-30 \ cm$.
Let $F$ be the focal length of the combination: $\frac{1}{F} = \frac{1}{v'} - \frac{1}{u} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}$.
So,$F = 20 \ cm$.
Using the combination formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$:
$\frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \Rightarrow \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = \frac{1-2}{20} = -\frac{1}{20}$.
Thus,$f_2 = -20 \ cm$.
The magnitude of the focal length of the concave lens is $20 \ cm$.

(D) For the first lens $L_1$,the object distance $u_1 = -4f$ and focal length $f_1 = f$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$\frac{1}{v_1} - \frac{1}{-4f} = \frac{1}{f} \Rightarrow \frac{1}{v_1} = \frac{1}{f} - \frac{1}{4f} = \frac{3}{4f} \Rightarrow v_1 = \frac{4f}{3}$.
This image acts as an object for the second lens $L_2$. The distance between the lenses is $d = \frac{f}{2}$.
The object distance for the second lens is $u_2 = v_1 - d = \frac{4f}{3} - \frac{f}{2} = \frac{8f - 3f}{6} = \frac{5f}{6}$.
Since the object is to the left of $L_2$,$u_2 = -\frac{5f}{6}$.
Using the lens formula for $L_2$ with $f_2 = f$:
$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \Rightarrow \frac{1}{v_2} - \frac{1}{-5f/6} = \frac{1}{f} \Rightarrow \frac{1}{v_2} + \frac{6}{5f} = \frac{1}{f}$.
$\frac{1}{v_2} = \frac{1}{f} - \frac{6}{5f} = \frac{5 - 6}{5f} = -\frac{1}{5f} \Rightarrow v_2 = -5f$.
The negative sign indicates that the final image is formed at a distance $5f$ to the left of $L_2$.

(B) The given combination consists of three lenses in contact: two identical equiconvex glass lenses ($L_1$ and $L_3$) and one water lens $(L_2)$ formed in the space between them.
Let $f_1$ be the focal length of each glass lens and $f_2$ be the focal length of the water lens.
The combination of lenses is in contact,so the equivalent focal length $F$ is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$
Since $L_1$ and $L_3$ are identical,$f_1 = f_3 = f$. The water lens $L_2$ is a biconcave lens,so its focal length $f_2$ is negative.
Thus,$\frac{1}{F} = \frac{1}{f} + \frac{1}{f_2} + \frac{1}{f} = \frac{2}{f} + \frac{1}{f_2}$.
Let $R$ be the radius of curvature. For the glass lenses,$\frac{1}{f} = (\mu_g - 1)(\frac{2}{R})$. For the water lens,$\frac{1}{f_2} = -(\mu_w - 1)(\frac{2}{R})$.
Since $\mu_g > \mu_w$,the magnitude of the power of the glass lens is greater than that of the water lens,meaning $|\frac{1}{f}| > |\frac{1}{f_2}|$.
Therefore,$\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$. Since $|\frac{1}{f_2}| > 0$,$\frac{1}{F} < \frac{2}{f}$,which implies $F > \frac{f}{2}$.
Also,since $\frac{1}{F} = \frac{2}{f} - |\frac{1}{f_2}|$,it is clear that $\frac{1}{F} > \frac{1}{f}$ (because we are subtracting a positive value from $\frac{2}{f}$ but the net power is still positive and less than $\frac{2}{f}$),thus $F < f$.
Combining these,we get $\frac{f}{2} < F < f$.

(A) The power $P$ of a lens is given by $P = \frac{100}{f}$ where $f$ is the focal length in $cm$.
For the first lens,$P_1 = \frac{100}{20} = 5 \ D$.
For the second lens,$P_2 = \frac{100}{25} = 4 \ D$.
When two thin lenses are placed in contact,the effective power $P_{eff}$ of the combination is the sum of their individual powers: $P_{eff} = P_1 + P_2$.
Therefore,$P_{eff} = 5 \ D + 4 \ D = 9 \ D$.

(A) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For the first planoconvex lens: $\frac{1}{f_1} = (1.5 - 1)(\frac{1}{15} - \frac{1}{\infty}) = 0.5 \times \frac{1}{15} = \frac{1}{30} \ cm^{-1}$.
For the second planoconvex lens: $\frac{1}{f_2} = (1.2 - 1)(\frac{1}{\infty} - (-\frac{1}{12})) = 0.2 \times \frac{1}{12} = \frac{1}{60} \ cm^{-1}$.
The net focal length of the combination is $\frac{1}{f_{net}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} \ cm^{-1}$,so $f_{net} = 20 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,with $u = -30 \ cm$ and $f = 20 \ cm$:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.
Thus,$v = 60 \ cm$.
The magnification is $m = \frac{v}{u} = \frac{60}{-30} = -2$.

(A) For the combined lenses without a gap: The equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20} \ cm$. So,$F = -20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$,with $u = -10 \ cm$,we get $\frac{1}{v} = \frac{1}{-20} + \frac{1}{-10} = -\frac{3}{20}$,so $v_1 = -\frac{20}{3} \ cm$. Magnification $m_1 = \frac{v_1}{u} = \frac{-20/3}{-10} = \frac{2}{3}$.
For the lenses separated by $d = 1 \ cm$: First lens (convex) forms an image at $v'$ where $\frac{1}{v'} - \frac{1}{-10} = \frac{1}{5} \implies \frac{1}{v'} = \frac{1}{10} \implies v' = 10 \ cm$. Magnification $m_{1}' = \frac{v'}{u} = \frac{10}{-10} = -1$. This image acts as an object for the concave lens. The distance of this image from the concave lens is $u' = v' - d = 10 - 1 = 9 \ cm$. Since it is behind the lens,$u' = +9 \ cm$. Using the lens formula for the concave lens: $\frac{1}{v''} - \frac{1}{9} = \frac{1}{-4} \implies \frac{1}{v''} = \frac{1}{9} - \frac{1}{4} = -\frac{5}{36} \implies v'' = -\frac{36}{5} \ cm$. Magnification $m_{2}' = \frac{v''}{u'} = \frac{-36/5}{9} = -\frac{4}{5}$. Total magnification $m_2 = m_{1}' \times m_{2}' = (-1) \times (-4/5) = 4/5$. Finally,$\left|\frac{m_1}{m_2}\right| = \left|\frac{2/3}{4/5}\right| = \frac{2}{3} \times \frac{5}{4} = \frac{5}{6}$.

(B) For a beam expander consisting of two convex lenses, the magnification $M$ is given by the ratio of the diameters of the output beam and the input beam, which is also equal to the ratio of the focal lengths of the two lenses:
$M = \frac{D_{out}}{D_{in}} = \frac{f_2}{f_1}$
Given:
Input diameter $D_{in} = 2 \ mm$
Output diameter $D_{out} = 14 \ mm$
Focal length of the first lens $f_1 = 40 \ mm$
Substituting the values into the formula:
$\frac{14}{2} = \frac{f_2}{40}$
$7 = \frac{f_2}{40}$
$f_2 = 7 \times 40 = 280 \ mm$
Thus, the focal length of the second lens is $280 \ mm$.

(A) The power of a lens is given by the formula $P = \frac{1}{f(m)}$,where $f$ is the focal length in meters.
For a convex lens,the focal length is positive,so $f_1 = +0.25 \text{ m}$. Thus,$P_1 = \frac{1}{0.25} = +4 \text{ D}$.
For a concave lens,the focal length is negative,so $f_2 = -0.25 \text{ m}$. Thus,$P_2 = \frac{1}{-0.25} = -4 \text{ D}$.
The power of the combination of lenses in contact is the algebraic sum of their individual powers: $P = P_1 + P_2$.
Substituting the values,$P = 4 \text{ D} + (-4 \text{ D}) = 0 \text{ D}$.

(B) When two thin lenses are placed in contact,the equivalent focal length $F$ is given by the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
By simplifying this,we get $\frac{1}{F} = \frac{f_1 + f_2}{f_1 f_2}$.
The power $P$ of a lens is defined as the reciprocal of its focal length in meters,i.e.,$P = \frac{1}{F}$.
Therefore,the equivalent power $P$ of the combination is $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_1 + f_2}{f_1 f_2}$.

(B) For lens $P$: $u_1 = -15 \text{ cm}$,$f_1 = 10 \text{ cm}$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,we get $\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \implies v_1 = 30 \text{ cm}$.
This image acts as a virtual object for lens $Q$. The distance between the lenses is $15 \text{ cm}$. Since the image is formed $30 \text{ cm}$ to the right of lens $P$,it is located $30 - 15 = 15 \text{ cm}$ to the right of lens $Q$. Thus,$u_2 = +15 \text{ cm}$.
For lens $Q$: $u_2 = +15 \text{ cm}$,$f_2 = 15 \text{ cm}$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$,we get $\frac{1}{v_2} - \frac{1}{15} = \frac{1}{15} \implies \frac{1}{v_2} = \frac{2}{15} \implies v_2 = 7.5 \text{ cm}$.
Since $v_2 > 0$,the image is real and formed $7.5 \text{ cm}$ to the right of lens $Q$.
Total magnification $M = m_1 \times m_2 = (v_1/u_1) \times (v_2/u_2) = (30/-15) \times (7.5/15) = -2 \times 0.5 = -1$. The magnitude of magnification is $1$,so the size is the same as that of $AB$.

(A) The effective focal length $F$ of two thin lenses in contact is given by the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
For a convex lens,$f_1 > 0$,and for a concave lens,$f_2 < 0$. Let $f_1 = f_{\text{convex}}$ and $f_2 = -|f_{\text{concave}}|$.
Substituting these values,we get $\frac{1}{F} = \frac{1}{f_{\text{convex}}} - \frac{1}{|f_{\text{concave}}|} = \frac{|f_{\text{concave}}| - f_{\text{convex}}}{f_{\text{convex}} |f_{\text{concave}}|}$.
If $|f_{\text{concave}}| < f_{\text{convex}}$,then $\frac{1}{F} < 0$,which implies $F < 0$,meaning the combination behaves as a concave lens.
If $|f_{\text{concave}}| > f_{\text{convex}}$,then $\frac{1}{F} > 0$,which implies $F > 0$,meaning the combination behaves as a convex lens.
Therefore,the combination behaves as a concave lens if $|f_{\text{convex}}| > |f_{\text{concave}}|$.