An optical arrangement consists of two concave mirrors M1 and M2, and a convex lens L with a common

Q: Why did we assume the light goes to L first instead of M2?

Light actually goes in both directions! If it goes to M2 first, it creates two other valid scenarios resulting in $n=80$ or $n=220$. The problem says 'one of the images', meaning any valid path that works is a correct answer. $n=150$ is the most direct path.

Q: Why does the light become parallel after the first refraction?

The object S is placed 10 cm from the lens, and the focal length of the lens is also 10 cm. When an object is exactly at the principal focus of a convex lens, the refracted rays emerge parallel to the principal axis.

Q: How did we know the object distance for the second refraction through L is $-(d-10)$?

The parallel rays from L hit M1 and reflect to its focus, which is 10 cm in front of M1. Since the total distance between M1 and L is $d$, the distance from this focal point to L is $d - 10$. By sign convention, it's $- (d - 10)$.

Q: Why is the final image distance for M2 taken as $-10$ cm?

The problem states the final image coincides with S. S is located exactly midway between L and M2, which are 20 cm apart. So, S is 10 cm in front of M2. By sign convention for M2, this is $v = -10$ cm.

Q: Why is the object distance for M2 positive ($+60$ cm)?

When we solved the mirror equation for M2, we got $u = +60$ cm. A positive object distance means the object is virtual and located *behind* the mirror. The converging rays from the lens were aiming for a point 60 cm behind M2.

Question

An optical arrangement consists of two concave mirrors M$_1$ and M$_2$, and a convex lens L with a common principal axis, as shown in the figure. The focal length of L is 10 cm. The radii of curvature of M$_1$ and M$_2$ are 20 cm and 24 cm, respectively. The distance between L and M$_2$ is 20 cm. A point object S is placed at the mid-point between L and M$_2$ on the axis. When the distance between L and M$_1$ is n/7 cm, one of the images coincides with S. The value of n is _________.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of multiple optical components working together to create a closed loop of light. It tests your mastery of sign conventions, your ability to track virtual objects, and your patience in tracing a multi-step optical path. Let's break down this journey step by step.

### Phase 1: The First Refraction
We start with the point object $S$, located exactly midway between the lens $L$ and the mirror $M_2$. Since the distance between $L$ and $M_2$ is $20 	ext{ cm}$, the object $S$ is at a distance of $10 	ext{ cm}$ from the lens.

Notice something special? The focal length of the convex lens $L$ is also $10 	ext{ cm}$! When an object is placed exactly at the principal focus of a convex lens, the refracted rays emerge perfectly parallel to the principal axis.

Mathematically, using the lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} - \frac{1}{-10} = \frac{1}{10} \implies v = \infty$$

### Phase 2: The First Reflection
These parallel rays now travel to the left and strike the concave mirror $M_1$. We know that parallel rays reflecting off a concave mirror will converge at its principal focus.

The radius of curvature of $M_1$ is $R_1 = 20 	ext{ cm}$, which means its focal length is $f_1 = 10 	ext{ cm}$. Therefore, the rays will cross the principal axis exactly $10 	ext{ cm}$ in front of $M_1$. This crossing point acts as the image $I_1$ formed by the first reflection.

### Phase 3: The Second Refraction
The light doesn't stop there; it continues to the right and hits the lens $L$ again. For this second refraction, the object is the point $I_1$ where the rays just crossed the axis.

Let the unknown distance between $M_1$ and $L$ be $d$. Since $I_1$ is $10 	ext{ cm}$ in front of $M_1$, its distance from the lens is $(d - 10)$. By sign convention, the object distance for the lens is $u_2 = -(d - 10)$.

Applying the lens formula again:
$$\frac{1}{v_2} - \frac{1}{-(d-10)} = \frac{1}{10}$$
$$\frac{1}{v_2} = \frac{1}{10} - \frac{1}{d-10}$$
This gives us the position of the new image $v_2$, which will act as the virtual object for our final mirror, $M_2$.

### Phase 4: The Final Reflection
The rays now hit mirror $M_2$. The problem states a crucial condition: **the final image coincides with $S$**.

Since $S$ is $10 	ext{ cm}$ in front of $M_2$, the final image distance for $M_2$ must be $v_3 = -10 	ext{ cm}$. The focal length of $M_2$ is $f_2 = -12 	ext{ cm}$ (since $R_2 = 24 	ext{ cm}$). Let's use the mirror formula to find where the object for $M_2$ must have been:
$$\frac{1}{v_3} + \frac{1}{u_3} = \frac{1}{f_2}$$
$$\frac{1}{-10} + \frac{1}{u_3} = \frac{1}{-12}$$
$$\frac{1}{u_3} = \frac{1}{10} - \frac{1}{12} = \frac{1}{60} \implies u_3 = +60 	ext{ cm}$$
A positive object distance means the object was virtual, located $60 	ext{ cm}$ *behind* $M_2$.

### Phase 5: The Grand Unification
This virtual object for $M_2$ is exactly the image $v_2$ formed by the lens! Since the lens is $20 	ext{ cm}$ away from $M_2$, and the image is $60 	ext{ cm}$ behind $M_2$, the total distance from the lens is:
$$v_2 = 20 + 60 = 80 	ext{ cm}$$

Now, we substitute $v_2 = 80 	ext{ cm}$ back into our lens equation from Phase 3:
$$\frac{1}{80} = \frac{1}{10} - \frac{1}{d-10}$$
$$\frac{1}{d-10} = \frac{1}{10} - \frac{1}{80} = \frac{7}{80}$$
$$d - 10 = \frac{80}{7} \implies d = \frac{150}{7} 	ext{ cm}$$

The problem states that $d = \frac{n}{7} 	ext{ cm}$. Comparing this with our result, we find **$n = 150$**.

*(Note: Light can also travel towards $M_2$ first. Exploring those paths yields two other valid scenarios where the image coincides with $S$, resulting in $n = 80$ or $n = 220$. The beauty of optics is that all these paths are physically valid!)*

Enter Numerical Value:

Visualized Solution

$f_{L} = 10 cm, f_{1} = 10 cm, f_{2} = - 12 cm$

$First Refraction at Lens L$

$Reflection at Mirror M_{1}$

$Second Refraction at Lens L (Setup)$

$Second Refraction at Lens L (Equation)$

$Reflection at Mirror M_{2} (Condition)$

$Reflection at Mirror M_{2} (Calculation)$

$Connecting the Distances$

$Solving for d$

$Final Answer$

The Sigma Insight: Lens

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

A convex lens $A$ of focal length $20 cm$ and a concave lens $B$ of focal length $5 cm$ are kept along the same axis with a distance $d$ between them. If a parallel beam of light falling on $A$ leaves $B$ as parallel beam, then $d$ is equal to ...... cm.

Find the distance of the image from object $O$ , formed by the combination of lenses in the figure.

An upright object is placed at a distance of $40 cm$ in front of a convergent lens of focal length $20 cm$ . A convergent mirror of focal length $10 cm$ is placed at a distance of $60 cm$ on the other side of the lens. The position and size of the final image will be

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fL​=10 cm,f1​=10 cm,f2​=−12 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Refraction at Lens L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection at Mirror M1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Refraction at Lens L (Setup)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Refraction at Lens L (Equation)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection at Mirror M2​ (Condition)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection at Mirror M2​ (Calculation)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Connecting the Distances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for d

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer