Comprehension
Read the following paragraph and answer the questions that follow.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.
Question: 1
The value of d is
The value of d is
Show Hint
Always remember that for the displacement method to work and provide real roots, the discriminant must be non-negative ($D^2 - 4fD \ge 0$).
This leads to the fundamental condition $D \ge 4f$ for a real image to form on the screen in two positions.
If $D = 4f$, the two positions coincide ($d = 0$), and if $D < 4f$, no sharp image can be formed on the screen.
This leads to the fundamental condition $D \ge 4f$ for a real image to form on the screen in two positions.
If $D = 4f$, the two positions coincide ($d = 0$), and if $D < 4f$, no sharp image can be formed on the screen.
Updated On: Sep 14, 2026
- $\sqrt{D(D - 4f)}$
- $\sqrt{D(D - 2f)}$
- $2\sqrt{Df}$
- $\sqrt{D(D - f)}$
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The Correct Option is A
Solution and Explanation
Concept:
• The question is based on the well-known displacement method used in optics to determine the focal length of a convex lens in laboratory experiments.
• When the distance between an object and a screen (denoted as $D$) is greater than or equal to $4f$, there exist exactly two positions of the convex lens for which a sharp real image is formed on the screen.
• This happens because of the principle of reversibility of light, which implies that the object and image distances can be interchanged (conjugate foci).
Step 1: Setting up the lens equation
Let the object be placed at the origin, and the screen is at a distance $D$.
Let the lens be placed at a distance $x$ from the object.
According to the sign convention, the object distance is $u = -x$.
Since the image forms on the screen, the image distance is $v = +(D - x)$.
The thin lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Step 2: Formulating the quadratic equation
Substitute the values of $u$ and $v$ into the lens formula:
\[ \frac{1}{D - x} - \frac{1}{-x} = \frac{1}{f} \]
\[ \frac{1}{D - x} + \frac{1}{x} = \frac{1}{f} \]
Taking the least common multiple (LCM) on the left-hand side:
\[ \frac{x + (D - x)}{x(D - x)} = \frac{1}{f} \]
\[ \frac{D}{Dx - x^2} = \frac{1}{f} \]
Cross-multiplying to rearrange into a standard quadratic equation form:
\[ Dx - x^2 = fD \]
\[ x^2 - Dx + fD = 0 \]
Step 3: Solving for the roots and finding their difference
The above equation is a quadratic equation in $x$, which means it will have two solutions (say $x_1$ and $x_2$) representing the two lens positions.
Using the quadratic formula, the roots are:
\[ x = \frac{D \pm \sqrt{D^2 - 4fD}}{2} \]
The two positions of the lens are $x_1 = \frac{D - \sqrt{D^2 - 4fD}}{2}$ and $x_2 = \frac{D + \sqrt{D^2 - 4fD}}{2}$.
The distance between these two lens positions is given as $d$, which is the difference between the roots:
\[ d = x_2 - x_1 \]
\[ d = \frac{D + \sqrt{D^2 - 4fD}}{2} - \frac{D - \sqrt{D^2 - 4fD}}{2} \]
\[ d = \frac{2\sqrt{D^2 - 4fD}}{2} = \sqrt{D^2 - 4fD} \]
Taking $D$ completely common inside the square root to match the options:
\[ d = \sqrt{D(D - 4f)} \]
Step 4: Conclusion
The derived expression for the separation between the two lens positions is mathematically $\sqrt{D(D - 4f)}$.
This exactly corresponds to option (A).
• The question is based on the well-known displacement method used in optics to determine the focal length of a convex lens in laboratory experiments.
• When the distance between an object and a screen (denoted as $D$) is greater than or equal to $4f$, there exist exactly two positions of the convex lens for which a sharp real image is formed on the screen.
• This happens because of the principle of reversibility of light, which implies that the object and image distances can be interchanged (conjugate foci).
Step 1: Setting up the lens equation
Let the object be placed at the origin, and the screen is at a distance $D$.
Let the lens be placed at a distance $x$ from the object.
According to the sign convention, the object distance is $u = -x$.
Since the image forms on the screen, the image distance is $v = +(D - x)$.
The thin lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Step 2: Formulating the quadratic equation
Substitute the values of $u$ and $v$ into the lens formula:
\[ \frac{1}{D - x} - \frac{1}{-x} = \frac{1}{f} \]
\[ \frac{1}{D - x} + \frac{1}{x} = \frac{1}{f} \]
Taking the least common multiple (LCM) on the left-hand side:
\[ \frac{x + (D - x)}{x(D - x)} = \frac{1}{f} \]
\[ \frac{D}{Dx - x^2} = \frac{1}{f} \]
Cross-multiplying to rearrange into a standard quadratic equation form:
\[ Dx - x^2 = fD \]
\[ x^2 - Dx + fD = 0 \]
Step 3: Solving for the roots and finding their difference
The above equation is a quadratic equation in $x$, which means it will have two solutions (say $x_1$ and $x_2$) representing the two lens positions.
Using the quadratic formula, the roots are:
\[ x = \frac{D \pm \sqrt{D^2 - 4fD}}{2} \]
The two positions of the lens are $x_1 = \frac{D - \sqrt{D^2 - 4fD}}{2}$ and $x_2 = \frac{D + \sqrt{D^2 - 4fD}}{2}$.
The distance between these two lens positions is given as $d$, which is the difference between the roots:
\[ d = x_2 - x_1 \]
\[ d = \frac{D + \sqrt{D^2 - 4fD}}{2} - \frac{D - \sqrt{D^2 - 4fD}}{2} \]
\[ d = \frac{2\sqrt{D^2 - 4fD}}{2} = \sqrt{D^2 - 4fD} \]
Taking $D$ completely common inside the square root to match the options:
\[ d = \sqrt{D(D - 4f)} \]
Step 4: Conclusion
The derived expression for the separation between the two lens positions is mathematically $\sqrt{D(D - 4f)}$.
This exactly corresponds to option (A).
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Question: 2
Compared to the size of the object, the images formed in the two positions of the lens are respectively
Compared to the size of the object, the images formed in the two positions of the lens are respectively
Show Hint
A highly useful relationship in the displacement method is that the product of the two magnifications is exactly unity ($m_1 \times m_2 = 1$).
Furthermore, if $O$ is the true size of the object and $I_1, I_2$ are the sizes of the two images, the actual size of the object is the geometric mean of the image sizes: $O = \sqrt{I_1 \times I_2}$.
Furthermore, if $O$ is the true size of the object and $I_1, I_2$ are the sizes of the two images, the actual size of the object is the geometric mean of the image sizes: $O = \sqrt{I_1 \times I_2}$.
Updated On: Sep 14, 2026
- reduced, enlarged
- reduced, reduced
- enlarged, enlarged
- enlarged, reduced
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The Correct Option is D
Solution and Explanation
Concept:
• The magnification $m$ of a thin lens is defined as the ratio of the height of the image to the height of the object.
• In terms of object distance $u$ and image distance $v$, the linear magnification is given by the formula $m = \frac{v}{u}$.
• When the absolute value of magnification $|m| > 1$, the image is strictly enlarged.
• When the absolute value of magnification $|m| < 1$, the image is strictly reduced or diminished.
• In the displacement method, the two lens positions are completely conjugate to each other, meaning $u_1 = v_2$ and $v_1 = u_2$.
Step 1: Analyzing the first lens position
When the student starts moving the lens away from the object towards the screen, the first position encountered is closer to the object.
Let the object distance be $u_1$ and the image distance (distance to the screen) be $v_1$.
Because the lens is physically closer to the object, we definitely have $|u_1| < |v_1|$.
The magnification for this first position is $m_1 = \frac{v_1}{u_1}$.
Since the numerator is geometrically larger than the denominator ($|v_1| > |u_1|$), the absolute value $|m_1|$ is greater than 1.
Therefore, the image formed at the first position is highly enlarged compared to the original object.
Step 2: Analyzing the second lens position
As the student continues to move the lens further towards the screen, the second position is reached.
At this new position, the lens is now physically closer to the screen and much farther from the object.
Let the new object distance be $u_2$ and the new image distance be $v_2$.
Due to the principle of reversibility (conjugate property), $u_2 = v_1$ and $v_2 = u_1$.
Since we already established $|v_1| > |u_1|$, it strictly means that $|u_2| > |v_2|$.
The magnification for this second position is $m_2 = \frac{v_2}{u_2}$.
Since the numerator is now geometrically smaller than the denominator ($|v_2| < |u_2|$), the absolute value $|m_2|$ is less than 1.
Therefore, the image formed at the second position is significantly reduced (diminished) compared to the original object.
Step 3: Conclusion
Summarizing the sequence of observations as the lens moves away from the object:
The first position yields an enlarged image.
The second position yields a reduced image.
Hence, the sequence of images formed is "enlarged, reduced", which matches option (D).
• The magnification $m$ of a thin lens is defined as the ratio of the height of the image to the height of the object.
• In terms of object distance $u$ and image distance $v$, the linear magnification is given by the formula $m = \frac{v}{u}$.
• When the absolute value of magnification $|m| > 1$, the image is strictly enlarged.
• When the absolute value of magnification $|m| < 1$, the image is strictly reduced or diminished.
• In the displacement method, the two lens positions are completely conjugate to each other, meaning $u_1 = v_2$ and $v_1 = u_2$.
Step 1: Analyzing the first lens position
When the student starts moving the lens away from the object towards the screen, the first position encountered is closer to the object.
Let the object distance be $u_1$ and the image distance (distance to the screen) be $v_1$.
Because the lens is physically closer to the object, we definitely have $|u_1| < |v_1|$.
The magnification for this first position is $m_1 = \frac{v_1}{u_1}$.
Since the numerator is geometrically larger than the denominator ($|v_1| > |u_1|$), the absolute value $|m_1|$ is greater than 1.
Therefore, the image formed at the first position is highly enlarged compared to the original object.
Step 2: Analyzing the second lens position
As the student continues to move the lens further towards the screen, the second position is reached.
At this new position, the lens is now physically closer to the screen and much farther from the object.
Let the new object distance be $u_2$ and the new image distance be $v_2$.
Due to the principle of reversibility (conjugate property), $u_2 = v_1$ and $v_2 = u_1$.
Since we already established $|v_1| > |u_1|$, it strictly means that $|u_2| > |v_2|$.
The magnification for this second position is $m_2 = \frac{v_2}{u_2}$.
Since the numerator is now geometrically smaller than the denominator ($|v_2| < |u_2|$), the absolute value $|m_2|$ is less than 1.
Therefore, the image formed at the second position is significantly reduced (diminished) compared to the original object.
Step 3: Conclusion
Summarizing the sequence of observations as the lens moves away from the object:
The first position yields an enlarged image.
The second position yields a reduced image.
Hence, the sequence of images formed is "enlarged, reduced", which matches option (D).
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Question: 3
If the distance between object and screen is $80.00$ cm and the lens forms sharp images at two positions separated by $20.00$ cm., the focal length of convex lens is
If the distance between object and screen is $80.00$ cm and the lens forms sharp images at two positions separated by $20.00$ cm., the focal length of convex lens is
Show Hint
To prevent heavy multiplication errors during manual calculation, you can smartly use the algebraic identity $a^2 - b^2 = (a - b)(a + b)$ in the numerator.
For example, $(80^2 - 20^2) = (80 - 20)(80 + 20) = (60)(100) = 6000$.
This trick usually saves time and heavily minimizes calculation blunders during timed competitive exams.
For example, $(80^2 - 20^2) = (80 - 20)(80 + 20) = (60)(100) = 6000$.
This trick usually saves time and heavily minimizes calculation blunders during timed competitive exams.
Updated On: Sep 14, 2026
- $15.50$ cm
- $18.75$ cm
- $20.50$ cm
- $22.75$ cm
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
• This problem provides practical numerical values to apply the mathematical formula derived for the displacement method.
• The total separation between the object and the screen is represented by the capital letter $D$.
• The total separation between the two conjugate positions of the convex lens is represented by the small letter $d$.
• The focal length $f$ of the convex lens can be determined directly by rearranging the formula $d = \sqrt{D^2 - 4fD}$.
Step 1: Identify the given quantities
From the text of the problem, the distance between the object and the screen is given as:
$D = 80.00 \text{ cm}$.
The distance separating the two sharp image positions of the lens is given as:
$d = 20.00 \text{ cm}$.
Step 2: Rearrange the displacement formula
We previously established that the distance between the two lens positions is:
\[ d = \sqrt{D^2 - 4fD} \]
To isolate the focal length $f$, we first square both sides of the equation:
\[ d^2 = D^2 - 4fD \]
Next, we bring the term containing $f$ to the left side and $d^2$ to the right side:
\[ 4fD = D^2 - d^2 \]
Finally, we divide both sides entirely by $4D$ to explicitly solve for $f$:
\[ f = \frac{D^2 - d^2}{4D} \]
Step 3: Substitute values and calculate
Now, substitute the provided numerical values into the derived expression:
\[ f = \frac{(80.00)^2 - (20.00)^2}{4 \times 80.00} \]
Calculate the squares of the terms in the numerator:
\[ (80.00)^2 = 6400 \]
\[ (20.00)^2 = 400 \]
Calculate the denominator:
\[ 4 \times 80.00 = 320 \]
Substitute these computed values back into the fraction:
\[ f = \frac{6400 - 400}{320} \]
\[ f = \frac{6000}{320} \]
Simplify the fraction by completely dividing by $10$ and then evaluating:
\[ f = \frac{600}{32} \]
\[ f = 18.75 \text{ cm} \]
Step 4: Conclusion
The calculated focal length of the given convex lens is precisely $18.75 \text{ cm}$.
This matches option (B).
• This problem provides practical numerical values to apply the mathematical formula derived for the displacement method.
• The total separation between the object and the screen is represented by the capital letter $D$.
• The total separation between the two conjugate positions of the convex lens is represented by the small letter $d$.
• The focal length $f$ of the convex lens can be determined directly by rearranging the formula $d = \sqrt{D^2 - 4fD}$.
Step 1: Identify the given quantities
From the text of the problem, the distance between the object and the screen is given as:
$D = 80.00 \text{ cm}$.
The distance separating the two sharp image positions of the lens is given as:
$d = 20.00 \text{ cm}$.
Step 2: Rearrange the displacement formula
We previously established that the distance between the two lens positions is:
\[ d = \sqrt{D^2 - 4fD} \]
To isolate the focal length $f$, we first square both sides of the equation:
\[ d^2 = D^2 - 4fD \]
Next, we bring the term containing $f$ to the left side and $d^2$ to the right side:
\[ 4fD = D^2 - d^2 \]
Finally, we divide both sides entirely by $4D$ to explicitly solve for $f$:
\[ f = \frac{D^2 - d^2}{4D} \]
Step 3: Substitute values and calculate
Now, substitute the provided numerical values into the derived expression:
\[ f = \frac{(80.00)^2 - (20.00)^2}{4 \times 80.00} \]
Calculate the squares of the terms in the numerator:
\[ (80.00)^2 = 6400 \]
\[ (20.00)^2 = 400 \]
Calculate the denominator:
\[ 4 \times 80.00 = 320 \]
Substitute these computed values back into the fraction:
\[ f = \frac{6400 - 400}{320} \]
\[ f = \frac{6000}{320} \]
Simplify the fraction by completely dividing by $10$ and then evaluating:
\[ f = \frac{600}{32} \]
\[ f = 18.75 \text{ cm} \]
Step 4: Conclusion
The calculated focal length of the given convex lens is precisely $18.75 \text{ cm}$.
This matches option (B).
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Question: 4
Consider a convex lens of focal length 15 cms. For which of the following values of object-screen distance, two positions of the object can be found to obtain sharp image on the screen ?
Consider a convex lens of focal length 15 cms. For which of the following values of object-screen distance, two positions of the object can be found to obtain sharp image on the screen ?
Show Hint
A classic conceptual question often asked in vivas or exams is: "What happens exactly if $D = 4f$?"
In that extremely specific scenario, the discriminant becomes exactly zero, implying the two roots are identical ($x_1 = x_2 = 2f$).
The lens has only one valid position perfectly midway between the object and screen, and the magnification is exactly $-1$ (same size, inverted).
In that extremely specific scenario, the discriminant becomes exactly zero, implying the two roots are identical ($x_1 = x_2 = 2f$).
The lens has only one valid position perfectly midway between the object and screen, and the magnification is exactly $-1$ (same size, inverted).
Updated On: Sep 14, 2026
- 45 cm
- 50 cm
- 55 cm
- 65 cm
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The Correct Option is D
Solution and Explanation
Concept:
• The question explores the fundamental physical condition required to perform the displacement method successfully.
• For a convex lens to form a real image of an object on a fixed screen, the total distance between the object and the screen must not be arbitrarily small.
• Mathematically, this condition is heavily derived from the discriminant of the quadratic equation connecting lens position and focal length.
• The discriminant must be strictly non-negative ($D^2 - 4fD \ge 0$) to yield real physical positions for the lens.
Step 1: Define the mathematical constraint
As proven in earlier parts, the separation $d$ between the two lens positions is given by $d = \sqrt{D^2 - 4fD}$.
For $d$ to represent a real, measurable physical distance (or zero), the expression purely inside the square root must be greater than or equal to zero.
\[ D^2 - 4fD \ge 0 \]
Since the distance $D$ is a physically positive quantity ($D > 0$), we can safely divide the entire inequality by $D$:
\[ D - 4f \ge 0 \]
\[ D \ge 4f \]
This means the absolute minimum distance between an object and its real image formed by a convex lens is exactly $4f$.
Step 2: Apply the specific given values
The problem states that the focal length of the convex lens is $f = 15 \text{ cm}$.
Substitute this specific focal length into the derived constraint:
\[ D_{min} = 4 \times f \]
\[ D_{min} = 4 \times 15 \text{ cm} \]
\[ D_{min} = 60 \text{ cm} \]
Therefore, to find two distinct (or strictly coincident) positions of the lens that form a sharp image, the object-screen distance $D$ must be at least $60 \text{ cm}$.
Step 3: Evaluate the given options
We carefully check each option against our mathematically established condition $D \ge 60 \text{ cm}$:
(A) $D = 45 \text{ cm}$: Since $45 < 60$, this is purely impossible.
(B) $D = 50 \text{ cm}$: Since $50 < 60$, this is purely impossible.
(C) $D = 55 \text{ cm}$: Since $55 < 60$, this is purely impossible.
(D) $D = 65 \text{ cm}$: Since $65 > 60$, this fully satisfies the condition and allows for two distinct real lens positions.
Step 4: Conclusion
The only provided option that is mathematically greater than the minimum required distance of $60 \text{ cm}$ is $65 \text{ cm}$.
Thus, option (D) is the correct answer.
• The question explores the fundamental physical condition required to perform the displacement method successfully.
• For a convex lens to form a real image of an object on a fixed screen, the total distance between the object and the screen must not be arbitrarily small.
• Mathematically, this condition is heavily derived from the discriminant of the quadratic equation connecting lens position and focal length.
• The discriminant must be strictly non-negative ($D^2 - 4fD \ge 0$) to yield real physical positions for the lens.
Step 1: Define the mathematical constraint
As proven in earlier parts, the separation $d$ between the two lens positions is given by $d = \sqrt{D^2 - 4fD}$.
For $d$ to represent a real, measurable physical distance (or zero), the expression purely inside the square root must be greater than or equal to zero.
\[ D^2 - 4fD \ge 0 \]
Since the distance $D$ is a physically positive quantity ($D > 0$), we can safely divide the entire inequality by $D$:
\[ D - 4f \ge 0 \]
\[ D \ge 4f \]
This means the absolute minimum distance between an object and its real image formed by a convex lens is exactly $4f$.
Step 2: Apply the specific given values
The problem states that the focal length of the convex lens is $f = 15 \text{ cm}$.
Substitute this specific focal length into the derived constraint:
\[ D_{min} = 4 \times f \]
\[ D_{min} = 4 \times 15 \text{ cm} \]
\[ D_{min} = 60 \text{ cm} \]
Therefore, to find two distinct (or strictly coincident) positions of the lens that form a sharp image, the object-screen distance $D$ must be at least $60 \text{ cm}$.
Step 3: Evaluate the given options
We carefully check each option against our mathematically established condition $D \ge 60 \text{ cm}$:
(A) $D = 45 \text{ cm}$: Since $45 < 60$, this is purely impossible.
(B) $D = 50 \text{ cm}$: Since $50 < 60$, this is purely impossible.
(C) $D = 55 \text{ cm}$: Since $55 < 60$, this is purely impossible.
(D) $D = 65 \text{ cm}$: Since $65 > 60$, this fully satisfies the condition and allows for two distinct real lens positions.
Step 4: Conclusion
The only provided option that is mathematically greater than the minimum required distance of $60 \text{ cm}$ is $65 \text{ cm}$.
Thus, option (D) is the correct answer.
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Question: 5
A thin convex lens of focal length 10 cm and another thin lens of focal length 'f' are placed coaxially in contact. If the power of their combination is $\frac{10}{3}$ D, the value of 'f' is }
A thin convex lens of focal length 10 cm and another thin lens of focal length 'f' are placed coaxially in contact. If the power of their combination is $\frac{10}{3}$ D, the value of 'f' is }
Show Hint
Always double-check units when dealing comprehensively with optical power.
A highly common mistake is using focal length directly in centimeters in the formula $P = 1/f$, which yields drastically wrong power values.
Remember the alternative handy formula: $P(D) = \frac{100}{f(in cm)}$ to skip the tedious meter conversion step entirely.
A highly common mistake is using focal length directly in centimeters in the formula $P = 1/f$, which yields drastically wrong power values.
Remember the alternative handy formula: $P(D) = \frac{100}{f(in cm)}$ to skip the tedious meter conversion step entirely.
Updated On: Sep 14, 2026
- $-15$ cm
- $-10$ cm
- $-20$ cm
- $-30$ cm
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The Correct Option is A
Solution and Explanation
Concept:
• When two or more thin lenses are placed coaxially in strict physical contact, their individual powers algebraically add up to give the total equivalent power of the combination.
• The total power $P_{eq}$ is given by the formula $P_{eq} = P_1 + P_2 + ...$
• The power of a single lens in Diopters (D) is calculated precisely as the reciprocal of its focal length measured in meters ($P = \frac{1}{f(in m)}$).
• Proper sign convention must be stringently followed: convex lenses have positive focal length and power, whereas concave lenses have negative focal length and power.
Step 1: Calculate the power of the first lens
The first lens is a convex lens, so its focal length is strictly positive.
Given focal length $f_1 = +10 \text{ cm}$.
We must convert this dimension perfectly into meters to properly calculate power in Diopters:
\[ f_1 = \frac{+10}{100} \text{ m} = +0.1 \text{ m} \]
Now, calculate the absolute power $P_1$ of this first lens:
\[ P_1 = \frac{1}{f_1} = \frac{1}{+0.1} = +10 \text{ D} \]
Step 2: Set up the combination equation
The total equivalent power of the lens combination is given in the problem as:
\[ P_{eq} = +\frac{10}{3} \text{ D} \]
Using the fundamental additive property of lens powers in contact:
\[ P_{eq} = P_1 + P_2 \]
Substitute the known values deeply into the equation:
\[ \frac{10}{3} = 10 + P_2 \]
Step 3: Solve for the power and focal length of the unknown lens
Isolate $P_2$ algebraically:
\[ P_2 = \frac{10}{3} - 10 \]
Take a common denominator to subtract correctly:
\[ P_2 = \frac{10 - 30}{3} = -\frac{20}{3} \text{ D} \]
The strongly negative sign securely indicates that the second lens is fundamentally diverging (a concave lens).
Now, convert this resulting power back into a focal length $f_2$ in meters:
\[ f_2 = \frac{1}{P_2} = \frac{1}{-\frac{20}{3}} \text{ m} = -\frac{3}{20} \text{ m} \]
Finally, convert this physical dimension back into centimeters for the final answer matching the options:
\[ f_2 = -\frac{3}{20} \times 100 \text{ cm} \]
\[ f_2 = -3 \times 5 \text{ cm} = -15 \text{ cm} \]
Step 4: Conclusion
The focal length 'f' of the unknown second thin lens is heavily calculated to be $-15 \text{ cm}$.
This beautifully matches option (A).
• When two or more thin lenses are placed coaxially in strict physical contact, their individual powers algebraically add up to give the total equivalent power of the combination.
• The total power $P_{eq}$ is given by the formula $P_{eq} = P_1 + P_2 + ...$
• The power of a single lens in Diopters (D) is calculated precisely as the reciprocal of its focal length measured in meters ($P = \frac{1}{f(in m)}$).
• Proper sign convention must be stringently followed: convex lenses have positive focal length and power, whereas concave lenses have negative focal length and power.
Step 1: Calculate the power of the first lens
The first lens is a convex lens, so its focal length is strictly positive.
Given focal length $f_1 = +10 \text{ cm}$.
We must convert this dimension perfectly into meters to properly calculate power in Diopters:
\[ f_1 = \frac{+10}{100} \text{ m} = +0.1 \text{ m} \]
Now, calculate the absolute power $P_1$ of this first lens:
\[ P_1 = \frac{1}{f_1} = \frac{1}{+0.1} = +10 \text{ D} \]
Step 2: Set up the combination equation
The total equivalent power of the lens combination is given in the problem as:
\[ P_{eq} = +\frac{10}{3} \text{ D} \]
Using the fundamental additive property of lens powers in contact:
\[ P_{eq} = P_1 + P_2 \]
Substitute the known values deeply into the equation:
\[ \frac{10}{3} = 10 + P_2 \]
Step 3: Solve for the power and focal length of the unknown lens
Isolate $P_2$ algebraically:
\[ P_2 = \frac{10}{3} - 10 \]
Take a common denominator to subtract correctly:
\[ P_2 = \frac{10 - 30}{3} = -\frac{20}{3} \text{ D} \]
The strongly negative sign securely indicates that the second lens is fundamentally diverging (a concave lens).
Now, convert this resulting power back into a focal length $f_2$ in meters:
\[ f_2 = \frac{1}{P_2} = \frac{1}{-\frac{20}{3}} \text{ m} = -\frac{3}{20} \text{ m} \]
Finally, convert this physical dimension back into centimeters for the final answer matching the options:
\[ f_2 = -\frac{3}{20} \times 100 \text{ cm} \]
\[ f_2 = -3 \times 5 \text{ cm} = -15 \text{ cm} \]
Step 4: Conclusion
The focal length 'f' of the unknown second thin lens is heavily calculated to be $-15 \text{ cm}$.
This beautifully matches option (A).
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