Combination of Lenses Questions in English

(A) The focal length of a convex lens is $f_1 = +40 \; cm$ and the focal length of a concave lens is $f_2 = -25 \; cm$.
The power of a lens is given by $P = \frac{100}{f(cm)} \; D$.
The power of the convex lens is $P_1 = \frac{100}{40} = +2.5 \; D$.
The power of the concave lens is $P_2 = \frac{100}{-25} = -4.0 \; D$.
The power of the combination of thin lenses in contact is the algebraic sum of their individual powers:
$P = P_1 + P_2$
$P = 2.5 \; D + (-4.0 \; D) = -1.5 \; D$.

(D) The formula for the focal length of two thin lenses in contact is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given: $F = 80 \ cm$ and $f_1 = 20 \ cm$.
Substituting the values: $\frac{1}{80} = \frac{1}{20} + \frac{1}{f_2}$.
Rearranging for $f_2$: $\frac{1}{f_2} = \frac{1}{80} - \frac{1}{20} = \frac{1 - 4}{80} = -\frac{3}{80}$.
Thus,$f_2 = -\frac{80}{3} \ cm$.
The power $P$ of a lens in Diopters $(D)$ is given by $P = \frac{100}{f(cm)}$.
$P_2 = \frac{100}{-80/3} = -\frac{300}{80} = -3.75 \ D$.

(A) The power of a combination of thin lenses in contact is given by the algebraic sum of their individual powers: $P = P_1 + P_2$.
Given $P_1 = +12 \ D$ and $P_2 = -2 \ D$.
Therefore,the total power $P = 12 \ D + (-2 \ D) = 10 \ D$.
The focal length $F$ of the combination in centimeters is given by the formula $F = \frac{100}{P}$.
Substituting the value of $P$,we get $F = \frac{100}{10} = 10 \ cm$.

(C) The focal length of a convex lens is $f_1 = +20 \ cm$ and that of a concave lens is $f_2 = -20 \ cm$.
When 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 values: $\frac{1}{F} = \frac{1}{20} + \frac{1}{-20} = 0$.
This implies $F = \infty$.
$A$ lens combination with an infinite focal length acts as a plane glass slab.
For a plane glass slab,the magnification $m = +1$.
Therefore,the image formed is of the same size as the object and is erect.

(B) The focal length of a convex lens is $f_1 = +10 \ cm$.
The focal length of a concave lens of the same material and focal length is $f_2 = -10 \ cm$.
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 values:
$\frac{1}{F} = \frac{1}{10} + \frac{1}{-10} = \frac{1}{10} - \frac{1}{10} = 0$
Therefore,$\frac{1}{F} = 0$,which implies $F = \infty$.
Thus,the focal length of the combination is $Infinity$.

(C) The focal length of a convex lens is $f_1 = +84 \ cm$ and the focal length of a concave lens is $f_2 = -12 \ cm$.
For a combination of thin lenses 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 values:
$\frac{1}{F} = \frac{1}{84} + \frac{1}{-12} = \frac{1 - 7}{84} = \frac{-6}{84} = -\frac{1}{14} \ cm^{-1}$
Thus,$F = -14 \ cm$.
The power $P$ of the combination in diopters $(D)$ is given by $P = \frac{100}{F(cm)}$:
$P = \frac{100}{-14} = -\frac{50}{7} \ D$.

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

(B) The focal lengths of the two lenses are $f_1 = 20 \; cm = 0.2 \; m$ and $f_2 = 25 \; cm = 0.25 \; m$.
The power of the individual lenses are $P_1 = \frac{1}{f_1} = \frac{1}{0.2} = 5 \; D$ and $P_2 = \frac{1}{f_2} = \frac{1}{0.25} = 4 \; D$.
When two thin lenses are placed in contact,the effective power of the combination is given by $P = P_1 + P_2$.
Therefore,$P = 5 \; D + 4 \; D = 9 \; D$.

(A) The power of the combination of lenses is given by the sum of the individual powers: $P = P_1 + P_2$.
Given $P_1 = 6 \ D$ and $P_2 = -2 \ D$.
Therefore,the combined power $P = 6 \ D + (-2 \ D) = 4 \ D$.
The focal length $f$ of the combination is the reciprocal of the total power: $f = \frac{1}{P}$.
Substituting the value of $P$,we get $f = \frac{1}{4} \ m = 0.25 \ m$.

(B) When lenses are in contact,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Since the object is at infinity,the image is formed at the focus $F = 60 \ cm$. Thus,$\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60}$ ... $(i)$.
When the lenses are separated by $d = 10 \ cm$,the equivalent focal length $F'$ is given by $\frac{1}{F'} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$. The image shifts by $30 \ cm$ towards the combination,so the new image distance is $60 \ cm - 30 \ cm = 30 \ cm$. Thus,$\frac{1}{30} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{10}{f_1 f_2}$ ... $(ii)$.
Substituting $(i)$ into $(ii)$: $\frac{1}{30} = \frac{1}{60} - \frac{10}{f_1 f_2} \implies \frac{10}{f_1 f_2} = \frac{1}{60} - \frac{1}{30} = -\frac{1}{60} \implies f_1 f_2 = -600$.
From $(i)$,$\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{60} \implies f_1 + f_2 = \frac{-600}{60} = -10$.
Solving $x^2 - (f_1+f_2)x + f_1 f_2 = 0 \implies x^2 + 10x - 600 = 0 \implies (x+30)(x-20) = 0$. Thus,the focal lengths are $20 \ cm$ and $-30 \ cm$.

(C) For an achromatic combination,the condition is given by $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$.
Here,$\omega_1$ and $\omega_2$ are the dispersive powers of the two lenses,and $f_1$ and $f_2$ are their respective focal lengths.
Since the dispersive powers $\omega_1$ and $\omega_2$ are always positive,the focal lengths $f_1$ and $f_2$ must have opposite signs for the sum to be zero.
This implies that one lens must be convex (positive focal length) and the other must be concave (negative focal length).
Therefore,an achromatic combination is formed by joining $1$ convex lens and $1$ concave lens.

(D) When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact,the equivalent focal length $F$ of the combination is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Taking the common denominator,we get:
$\frac{1}{F} = \frac{f_2 + f_1}{f_1 f_2}$
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 power of the combination is $P = \frac{f_1 + f_2}{f_1 f_2}$.

(B) For an achromatic combination of two thin lenses in contact,the condition is that the net dispersive power of the combination must be zero.
The condition for achromatism is given by the equation: $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$.
Multiplying the entire equation by $f_1 f_2$,we get: $\omega_1 f_2 + \omega_2 f_1 = 0$.
Therefore,the correct condition is $\omega_1 f_2 + \omega_2 f_1 = 0$.

(A) The power $P$ of a lens is given by $P = \frac{1}{f}$,where $f$ is the focal length in meters.
For a convex lens,the focal length is positive,so $f_1 = 0.5 \ m$. Its power is $P_1 = \frac{1}{0.5} = 2 \ D$.
For a concave lens,the focal length is negative,so $f_2 = -1 \ m$. Its power is $P_2 = \frac{1}{-1} = -1 \ D$.
When lenses are combined,the total power is the sum of individual powers: $P = P_1 + P_2$.
$P = 2 \ D + (-1 \ D) = 1 \ D$.

(D) The power of a combination of thin lenses in contact is given by the algebraic sum of their individual powers: $P = P_1 + P_2$.
Given $P_1 = +2.50 \ D$ and $P_2 = -3.75 \ D$.
Therefore,the total power $P = 2.50 \ D + (-3.75 \ D) = -1.25 \ D$.
The focal length $f$ in centimeters is given by the formula $f = \frac{100}{P}$.
Substituting the value of $P$: $f = \frac{100}{-1.25} = -80 \ cm$.
Thus,the focal length of the combination is $-80 \ cm$.

(D) The power of a lens is given by $P = \frac{1}{f}$ (where $f$ is in meters).
For a convex lens,the focal length $f_1 = + 80 \; cm = + 0.8 \; m$.
For a concave lens,the focal length $f_2 = - 50 \; cm = - 0.5 \; m$.
The power of the combination is $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values: $P = \frac{1}{0.8} + \frac{1}{-0.5} = 1.25 - 2.0 = - 0.75 \; D$.

(A) When two thin lenses with powers $P_1$ and $P_2$ are placed in contact,the equivalent power $P$ of the combination is given by the algebraic sum of their individual powers.
$P = P_1 + P_2$
Given $P_1 = +2 \ D$ and $P_2 = -4 \ D$.
Substituting these values into the formula:
$P = (+2 \ D) + (-4 \ D) = -2 \ D$.
Therefore,the power of the combination is $-2 \ D$.

(B) When two thin lenses are placed in contact,the equivalent power $P$ of the combination is given by the algebraic sum of the individual powers: $P = P_1 + P_2$.
Given $P_1 = +6 D$ and $P_2 = -4 D$.
Therefore,$P = +6 D + (-4 D) = +2 D$.
Since the equivalent power $P$ is positive,the combination behaves as a convex lens.
The focal length $f$ is given by $f = \frac{1}{P}$ (in meters) or $f = \frac{100}{P}$ (in cm).
$f = \frac{100}{2} = +50 cm$.
Thus,the combination is a convex lens with a focal length of $50 cm$.

(D) The formula for the focal length of two thin lenses in contact is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given,the focal length of the convex lens is $f_1 = 9 \ cm$ and the focal length of the concave lens is $f_2 = -18 \ cm$.
Substituting these values into the formula:
$\frac{1}{F} = \frac{1}{9} + \frac{1}{-18}$
$\frac{1}{F} = \frac{2}{18} - \frac{1}{18}$
$\frac{1}{F} = \frac{1}{18}$
Therefore,the focal length of the combination is $F = 18 \ cm$.

(C) When two thin lenses are placed in contact, the equivalent power $P$ of the combination is the algebraic sum of the individual powers:
$P = P_1 + P_2$
Given $P_1 = +2 \text{ D}$ and $P_2 = -1 \text{ D}$, the equivalent power is:
$P = (+2 \text{ D}) + (-1 \text{ D}) = +1 \text{ D}$
Since the power is positive, the combination behaves as a convergent (convex) lens.
The focal length $f$ is given by:
$f = \frac{100}{P} \text{ cm} = \frac{100}{+1} \text{ cm} = +100 \text{ cm}$
Thus, the combination acts as a convergent lens of focal length $100 \text{ cm}$.

(A) For a combination of two thin lenses in contact,the equivalent focal length $F$ is given by the formula: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Here,$f_1$ is the focal length of the convex lens (positive,so $f_1 > 0$) and $f_2$ is the focal length of the concave lens (negative,so let $f_2 = -|f_2|$).
Substituting these,we get: $\frac{1}{F} = \frac{1}{f_1} - \frac{1}{|f_2|} = \frac{|f_2| - f_1}{f_1 |f_2|}$.
Thus,$F = \frac{f_1 |f_2|}{|f_2| - f_1}$.
For the combination to behave like a concave lens,the equivalent focal length $F$ must be negative.
This occurs when the denominator $|f_2| - f_1 < 0$,which implies $f_1 > |f_2|$.
Therefore,the combination behaves like a concave lens if $f_1 > f_2$ (where $f_2$ represents the magnitude of the focal length).

(A) The focal length of a convex lens is taken as positive,so $f_1 = +18 \ cm$.
The focal length of a concave lens is taken as negative,so $f_2 = -9 \ cm$.
The equivalent focal length $F$ of the combination is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting the values:
$\frac{1}{F} = \frac{1}{18} + \frac{1}{-9}$
$\frac{1}{F} = \frac{1}{18} - \frac{2}{18} = -\frac{1}{18}$
Therefore,$F = -18 \ cm$.
Since the equivalent focal length is negative,the combination acts as a concave lens with a focal length of $18 \ cm$.

(A) The power $P$ of a lens is given by the formula $P = \frac{100}{f}$ (where $f$ is in $cm$).
For a convex lens,the focal length $f_1 = +25 \ cm$. Therefore,its power $P_1 = \frac{100}{25} = +4 \ D$.
For a concave lens,the focal length $f_2 = -10 \ cm$. Therefore,its power $P_2 = \frac{100}{-10} = -10 \ D$.
The power of the combination of lenses in contact is given by $P_{eq} = P_1 + P_2$.
$P_{eq} = (+4 \ D) + (-10 \ D) = -6 \ D$.
Thus,the power of the combination is $-6 \ D$.

(C) For lens $1$: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f = +8 \ cm$ and $u = -12 \ cm$.
$\frac{1}{8} = \frac{1}{v_1} - \frac{1}{-12} \Rightarrow \frac{1}{v_1} = \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24}$.
So,$v_1 = 24 \ cm$. The image $A'B'$ is formed $24 \ cm$ to the right of the first lens.
The distance of this image from the second lens is $d = 30 \ cm - 24 \ cm = 6 \ cm$.
Since the focal length of the second lens is $f_2 = 6 \ cm$,the image $A'B'$ acts as an object placed exactly at the focus of the second lens.
Therefore,the rays emerging from the second lens will be parallel,and the final image will be formed at infinity. In optical terms,an image at infinity is considered real and highly enlarged.

(B) When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact coaxially,the equivalent focal length $F$ of the combination is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
To find $F$,we take the common denominator:
$\frac{1}{F} = \frac{f_2 + f_1}{f_1 f_2}$
Taking the reciprocal of both sides,we get:
$F = \frac{f_1 f_2}{f_1 + f_2}$
Therefore,the correct option is $B$.

(C) The focal length of a plano-convex lens is $f_1 = f$. The focal length of a thin concave lens is $f_2 = -f$ (since it is a concave lens of the same focal length).
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 values:
$\frac{1}{F} = \frac{1}{f} + \frac{1}{-f} = 0$
Therefore,$F = \infty$.
Since the equivalent focal length is infinite,the parallel rays will remain parallel after passing through the combination,meaning the focus shifts to infinity.

(B) For a combination of two thin lenses to have minimum spherical and chromatic aberration,the distance $d$ between the two lenses must satisfy the condition $d = f_1 - f_2$,where $f_1$ and $f_2$ are the focal lengths of the two lenses.
Given,$f_1 = 0.3 \, m$ and $f_2 = 0.1 \, m$.
Substituting these values into the formula:
$d = 0.3 \, m - 0.1 \, m = 0.2 \, m$.
Therefore,the required distance is $0.2 \, m$.

(B) Given: Total power of the combination $P = P_1 + P_2 = + 2D$.
Power of the convex lens $P_1 = + 5D$.
Therefore,the power of the concave lens $P_2 = P - P_1 = 2D - 5D = - 3D$.
For an achromatic combination,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which can be written in terms of power as $\omega_1 P_1 + \omega_2 P_2 = 0$.
This implies $\frac{\omega_1}{\omega_2} = - \frac{P_2}{P_1}$.
Substituting the values: $\frac{\omega_1}{\omega_2} = - \left( \frac{- 3D}{5D} \right) = \frac{3}{5}$.
Thus,the ratio of the dispersive power of the convex lens to the concave lens is $3:5$.

(D) The power of the combination of lenses in contact is given by the sum of their individual powers: $P = P_1 + P_2$.
Given $P_1 = +12 \ D$ and $P_2 = -2 \ D$.
Therefore,the total power $P = 12 \ D - 2 \ D = 10 \ D$.
The focal length $F$ in meters is given by $F = \frac{1}{P}$.
$F = \frac{1}{10} \ m = 0.1 \ m$.
To convert the focal length into centimeters,multiply by $100$: $F = 0.1 \times 100 \ cm = 10 \ cm$.

(D) Let the focal length of the convex lens be $f_1$ and the concave lens be $f_2$. Since the lens is concave,$f_2$ is negative.
Given the ratio of magnitudes: $|f_1| / |f_2| = 2/3$,so $f_1 / (-f_2) = 2/3$,which implies $f_2 = -1.5 f_1$.
The formula for the equivalent focal length $F$ of two thin lenses in contact is $1/F = 1/f_1 + 1/f_2$.
Given $F = 30 \ cm$,we have $1/30 = 1/f_1 - 1/|f_2|$.
Substituting $|f_2| = 1.5 f_1$: $1/30 = 1/f_1 - 1/(1.5 f_1) = (1.5 - 1) / (1.5 f_1) = 0.5 / (1.5 f_1) = 1 / (3 f_1)$.
Thus,$3 f_1 = 30$,so $f_1 = 10 \ cm$.
Then,$|f_2| = 1.5 \times 10 = 15 \ cm$. Since it is a concave lens,$f_2 = -15 \ cm$.
Therefore,the focal lengths are $10 \ cm$ and $-15 \ cm$.

(B) When two thin lenses are placed in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given $f_1 = f_2 = 40 \ cm$,the equivalent focal length is $\frac{1}{F} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}$,so $F = 20 \ cm$.
To obtain a real,inverted image of the same size (magnification $m = -1$),the object must be placed at a distance of $2F$ from the lens.
Therefore,the object distance $u = 2 \times 20 \ cm = 40 \ cm$.

(A) The equivalent power $P$ of two lenses separated by a distance $d$ is given by the formula: $P = P_1 + P_2 - d \cdot P_1 \cdot P_2$.
Given $P_1 = +5 \ D$ and $P_2 = +5 \ D$.
For the equivalent power to be negative,we require $P < 0$.
Substituting the values: $5 + 5 - d(5)(5) < 0$.
$10 - 25d < 0$.
$10 < 25d$.
$d > \frac{10}{25} \ m$.
$d > 0.4 \ m$.
Converting to centimeters: $d > 0.4 \times 100 \ cm = 40 \ cm$.
Thus,the equivalent power will be negative if the distance is greater than $40 \ cm$.

(C) The focal length of a convex lens is $f_1 = +20 \ cm$ and the focal length of a concave lens is $f_2 = -20 \ cm$.
When placed in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values: $\frac{1}{F} = \frac{1}{20} + \frac{1}{-20} = 0$,which implies $F = \infty$.
$A$ combination with infinite focal length acts as a plane glass plate.
$A$ plane glass plate forms a virtual,erect image of the same size as the object at the same distance behind the plate.

(B) For an achromatic combination,the condition is $\frac{f_1}{\omega_1} + \frac{f_2}{\omega_2} = 0$,which implies $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.
Given $\omega_1 = 0.024$ and $\omega_2 = 0.036$,we have $\frac{f_1}{f_2} = -\frac{0.024}{0.036} = -\frac{2}{3}$,so $f_2 = -1.5 f_1$.
The equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{90}$.
Substituting $f_2 = -1.5 f_1$ into the equation: $\frac{1}{f_1} - \frac{1}{1.5 f_1} = \frac{1}{90}$.
$\frac{1.5 - 1}{1.5 f_1} = \frac{1}{90} \implies \frac{0.5}{1.5 f_1} = \frac{1}{90} \implies \frac{1}{3 f_1} = \frac{1}{90}$.
Thus,$f_1 = 30 \ cm$ and $f_2 = -1.5(30) = -45 \ cm$.

(B) For a combination of two thin lenses made of the same material,the condition to minimize chromatic aberration is that the distance $d$ between the lenses must be equal to the average of their focal lengths.
Mathematically,this is expressed as:
$d = \frac{f_1 + f_2}{2}$
Therefore,the correct option is $B$.

(A) For a combination of two thin lenses to produce no net dispersion (achromatic doublet),the condition is given by $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$.
This implies $\frac{\omega_1}{\omega_2} = -\frac{f_1}{f_2}$.
Given the ratio of focal lengths is $\frac{f_1}{f_2} = \frac{2}{3}$.
Substituting this value,we get $\frac{\omega_1}{\omega_2} = -\frac{2}{3}$.
Since dispersive power is a positive quantity,we consider the magnitude of the ratio,which is $2:3$.

(B) The power $P$ of a lens is given by $P = \frac{100}{f(cm)}\, D$.
For the convex lens,the focal length $f_1 = +40\, cm$.
Therefore,the power $P_1 = \frac{100}{40} = +2.5\, D$.
For the concave lens,the focal length $f_2 = -25\, cm$.
Therefore,the power $P_2 = \frac{100}{-25} = -4.0\, D$.
The power of the combination of lenses in contact is given by $P = P_1 + P_2$.
$P = 2.5\, D + (-4.0\, D) = -1.5\, D$.

(B) $1$. The convex lens forms an image $I_1$ at its focal point. Since the object is at infinity, the image $I_1$ is formed at $f = 30\,cm$ from the convex lens.
$2$. The size of this image $I_1$ is $2\,cm$.
$3$. A concave lens of focal length $f_2 = -20\,cm$ is placed at $26\,cm$ from the convex lens. The image $I_1$ acts as a virtual object for the concave lens.
$4$. The distance of the virtual object from the concave lens is $u = 30\,cm - 26\,cm = 4\,cm$. Since it is on the right side, we take $u = +4\,cm$.
$5$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{4} = \frac{1}{-20}$
$\frac{1}{v} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$
$v = 5\,cm$.
$6$. The magnification $m$ for the concave lens is $m = \frac{v}{u} = \frac{5}{4} = 1.25$.
$7$. The new size of the image $I_2$ is $m \times (\text{size of } I_1) = 1.25 \times 2\,cm = 2.5\,cm$.

(C) Let the focal length of the convex lens be $f_1 = +30 \ cm$ and the concave lens be $f_2 = -10 \ cm$.
Let the distance between the two lenses be $d$.
For a parallel beam of light to remain parallel after passing through the combination,the effective focal length $F$ of the system must be infinite $(F = \infty)$.
The formula for the effective focal length of two lenses separated by a distance $d$ is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Substituting $F = \infty$,$f_1 = 30$,and $f_2 = -10$:
$0 = \frac{1}{30} + \frac{1}{-10} - \frac{d}{(30)(-10)}$
$0 = \frac{1}{30} - \frac{1}{10} + \frac{d}{300}$
Multiplying by $300$:
$0 = 10 - 30 + d$
$d = 20 \ cm$.
Thus,the distance between the two lenses is $20 \ cm$.

(A) The equivalent power of two lenses in contact is given by $P = P_1 + P_2$.
Substituting the given values: $P = 3.5 \, D + (-2.5 \, D) = 1.0 \, D$.
Since the power is positive,the combination acts as a converging lens.
The focal length $f$ is given by $f = \frac{100}{P} \, cm$.
$f = \frac{100}{1} = 100 \, cm$.
Therefore,the combination behaves as a converging lens of focal length $100 \, cm$.

(B) The power of the combination of lenses in contact is given by the sum of their individual powers: $P = P_1 + P_2$.
Given $P_1 = -15 \, D$ and $P_2 = 5 \, D$.
Therefore,$P = -15 \, D + 5 \, D = -10 \, D$.
The focal length $f$ of the combination is given by $f = \frac{1}{P}$ (in meters).
$f = \frac{1}{-10} \, m = -0.1 \, m$.
To convert the focal length into centimeters,multiply by $100$: $f = -0.1 \times 100 \, cm = -10 \, cm$.

(D) When a convex lens of focal length $f$ is cut perpendicular to its principal axis into two equal parts,each part becomes a plano-convex lens with a focal length of $2f$.
Given $f = 20 \, cm$,the focal length of each plano-convex lens is $f' = 2f = 40 \, cm$.
When these two lenses are placed in contact as shown in the figure (with their curved surfaces facing each other),the equivalent focal length $F$ is given by the lens combination formula:
$\frac{1}{F} = \frac{1}{f'} + \frac{1}{f'} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}$.
Therefore,$F = 20 \, cm$.

(B) When two thin lenses are placed in contact,the total power $P$ of the combination is the algebraic sum of the individual powers of the lenses.
Given: $P_1 = 6 \, D$ and $P_2 = -2 \, D$.
The total power $P = P_1 + P_2 = 6 \, D + (-2 \, D) = 4 \, D$.
The focal length $f$ of the combination is given by the formula $f = \frac{1}{P}$.
Substituting the value of $P$,we get $f = \frac{1}{4} \, m$.

(B) For an achromatic combination of two thin lenses in contact,the condition for zero chromatic aberration is given by:
$\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$
Rearranging the equation to find the ratio of the dispersive powers $\frac{\omega_1}{\omega_2}$:
$\frac{\omega_1}{\omega_2} = -\frac{f_1}{f_2}$
Given the focal lengths $f_1 = +10 \ cm$ and $f_2 = -15 \ cm$:
$\frac{\omega_1}{\omega_2} = -\frac{10}{-15}$
$\frac{\omega_1}{\omega_2} = \frac{10}{15} = \frac{2}{3}$
Thus,the ratio of their dispersive powers is $2/3$.

(C) The power of the combination of lenses is given by $P = P_1 + P_2$.
Given $P_1 = +3\,D$ and $P_2 = -5\,D$,the total power is $P = 3 + (-5) = -2\,D$.
The focal length $f$ of the combination is $f = \frac{1}{P} = \frac{1}{-2}\,m = -0.5\,m = -50\,cm$.
The object distance is $u = -50\,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we have $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$.
Substituting the values,$\frac{1}{v} = \frac{1}{-50} + \frac{1}{-50} = -\frac{2}{50} = -\frac{1}{25}$.
Therefore,$v = -25\,cm$.

(D) Let the focal length of the convex lens be $f_1 = 10 \, cm$ and the focal length of the concave lens be $f_2 = -f$ (where $f > 0$).
According to the lens combination formula,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given that the focal length of the combination is numerically equal to that of the concave lens,we have $F = -f$.
Substituting the values: $\frac{1}{-f} = \frac{1}{10} + \frac{1}{-f}$.
Wait,the problem states the combination's focal length is numerically equal to the concave lens. Let the concave lens focal length be $-x$. Then the combination focal length is $-x$.
So,$\frac{1}{-x} = \frac{1}{10} - \frac{1}{x}$.
Adding $\frac{1}{x}$ to both sides: $0 = \frac{1}{10}$. This implies a contradiction unless the combination focal length is defined differently.
Re-evaluating: If the combination focal length $F$ is numerically equal to the concave lens focal length $f_c$,and $f_c$ is negative,then $F = f_c$.
$\frac{1}{f_c} = \frac{1}{10} + \frac{1}{f_c}$. This is impossible.
However,if the magnitude is equal,$|F| = |f_c|$. Since the combination is $F = \frac{10(-f)}{10-f}$,and we set $F = f$ (magnitude),then $f = \frac{10f}{f-10} \Rightarrow f-10 = 10 \Rightarrow f = 20$.
Thus,the concave lens focal length is $-20 \, cm$.

(B) The total power of the combination is $P = P_1 + P_2 = +2 \text{ D}$.
Given the power of the convex lens $P_1 = +5 \text{ D}$, we find the power of the second lens $P_2 = P - P_1 = 2 - 5 = -3 \text{ D}$.
For an achromatic doublet, the condition is $\omega_1 P_1 + \omega_2 P_2 = 0$, where $\omega_1$ and $\omega_2$ are the dispersive powers.
This implies $\frac{\omega_1}{\omega_2} = -\frac{P_2}{P_1}$.
Substituting the values, $\frac{\omega_1}{\omega_2} = -\left( \frac{-3}{5} \right) = \frac{3}{5}$.
Thus, the ratio of the dispersive powers is $3 : 5$.

(A) When a convex lens of focal length $f$ is cut into two equal parts along its principal axis (vertically),each part becomes a plano-convex lens.
According to the lens maker's formula,$\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For the original lens,$R_1 = R$ and $R_2 = -R$,so $\frac{1}{f} = (n-1) \left( \frac{2}{R} \right)$.
When cut vertically,each new lens has one surface with radius $R$ and the other surface is flat $(R = \infty)$.
Thus,for the new lens,$\frac{1}{f'} = (n-1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n-1}{R}$.
Comparing the two,we get $f' = 2f$.
Since both parts are identical,their focal lengths are equal,so the ratio is $2f : 2f = 1 : 1$.

(C) When a biconvex lens is cut along its principal axis (horizontal cut),each half has the same focal length $f$ as the original lens.
This is because the radius of curvature of the surfaces remains unchanged for each part.
Thus,the power of each part remains $P = 1/f$.
When these two parts are joined as shown in figure $(b)$,they form a combination of two lenses in contact.
The total power of the combination is the sum of the powers of the individual lenses: $P_{net} = P_1 + P_2$.
Since $P_1 = P$ and $P_2 = P$,the net power is $P_{net} = P + P = 2P$.