Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, sep

Q: Why does the focal length of the mirror remain unchanged in the liquid?

The focal length of a spherical mirror depends purely on its geometry, specifically its radius of curvature ($f = R/2$). It does not depend on the refractive index of the surrounding medium, unlike a lens which relies on refraction.

Q: How do we determine the object distance for the lens?

The image formed by the mirror acts as the object for the lens. You must calculate the distance of this image from the mirror, and then use the given separation between the mirror and the lens (50 cm) to find its exact distance from the lens's optical center.

Q: Why do we multiply the magnifications?

Magnification is a scaling factor. The mirror scales the original object by a factor of $m_1$, creating an intermediate image. The lens then takes this intermediate image and scales it by a factor of $m_2$. The total scaling factor is the product of the two.

Q: What happens if the medium's refractive index is greater than the lens's refractive index?

If the surrounding medium is optically denser than the lens material ($\mu_m > \mu_g$), the term $(\frac{\mu_g}{\mu_m} - 1)$ becomes negative. This causes the focal length to change sign, meaning a convex lens would start behaving like a concave (diverging) lens!

Q: Why is the final image erect if both magnifications are negative?

A negative magnification implies an inverted image. The mirror inverts the original object (first negative sign), and the lens inverts that intermediate image again (second negative sign). Inverting an already inverted image makes it erect relative to the original object.

Question

Consider a concave mirror and a convex lens (refractive index $= 1.5$) of focal length $10	ext{ cm}$ each, separated by a distance of $50	ext{ cm}$ in air (refractive index $= 1$) as shown in the figure. An object is placed at a distance of $15	ext{ cm}$ from the mirror. Its erect image formed by this combination has magnification $M_1$.
When the set-up is kept in a medium of refractive index $7/6$, the magnification becomes $M_2$. The magnitude $\left|\frac{M_2}{M_1}ight|$ is

Sigma Percentile · Accepted Answer

### The Journey of Light: A Two-Part Symphony

Imagine you are a photon, embarking on a thrilling journey through an optical obstacle course. In this beautiful problem, our obstacle course consists of a concave mirror and a convex lens separated by a vast expanse of $50	ext{ cm}$. An object is placed $15	ext{ cm}$ in front of the mirror. The light will first reflect from the mirror, and then the reflected rays will travel towards the lens to form the final image. Let's break this journey into two distinct parts.

### Reflection at the Concave Mirror

The first encounter is with the concave mirror. The mirror formula is our trusty map:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Taking the pole of the mirror as the origin, the object is at $u_1 = -15	ext{ cm}$, and the focal length is $f_m = -10	ext{ cm}$. Substituting these values, we find:

$$\frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-10}$$

Solving this yields $v_1 = -30	ext{ cm}$. This means the mirror forms a real, inverted image $30	ext{ cm}$ in front of it.

Now, what about the magnification? The magnification produced by a mirror is $m = -\frac{v}{u}$. Plugging in our values, we get $m_1 = -\frac{-30}{-15} = -2$. The negative sign confirms the image is inverted, and it is twice the size of the object. This intermediate image, $I_1$, will now act as a virtual object for our convex lens.

### Refraction through the Convex Lens (in Air)

Next, the light rays, now converging to form the intermediate image, march towards the convex lens. Since the total distance between the mirror and the lens is $50	ext{ cm}$, and $I_1$ is $30	ext{ cm}$ from the mirror, it is exactly $20	ext{ cm}$ in front of the lens. So, for the lens, our object distance $u_2$ is $-20	ext{ cm}$.

The lens formula guides us here:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Substituting $u_2 = -20	ext{ cm}$ and the focal length $f_l = +10	ext{ cm}$, we get:

$$\frac{1}{v_2} - \frac{1}{-20} = \frac{1}{10}$$

Solving this, we find $v_2 = +20	ext{ cm}$. The lens forms the final image $20	ext{ cm}$ to its right. The magnification produced by the lens is $m_2 = \frac{v_2}{u_2} = \frac{20}{-20} = -1$.

The total magnification of the system, $M_1$, is the product of the individual magnifications:

$$M_1 = m_1 	imes m_2 = (-2) 	imes (-1) = 2$$

The final image is erect and twice the original size.

### The Plot Twist: Immersing the Setup in a Medium

But wait, a plot twist! The entire universe of our setup is suddenly submerged in a liquid of refractive index $\mu_m = \frac{7}{6}$.

The mirror, stoic and unyielding, doesn't care. Its focal length is locked in its geometry ($f = \frac{R}{2}$), so it remains $-10	ext{ cm}$. The intermediate image $I_1$ is still formed at the exact same spot, and $m_1$ remains $-2$.

The lens, however, is a creature of refraction. It feels the change in the surrounding medium. The Lens Maker's formula is the key to unlocking its new identity:

$$\frac{1}{f} = \left(\frac{\mu_g}{\mu_m} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$

First, let's find the geometric constant $K = \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$ of the lens using its focal length in air.

$$\frac{1}{10} = \left(\frac{1.5}{1} - 1ight) K \Rightarrow K = \frac{1}{5}$$

Now, in the new medium, the relative refractive index is $\frac{1.5}{7/6} = \frac{9}{7}$. Substituting this back into the Lens Maker's formula:

$$\frac{1}{f_{med}} = \left(\frac{9}{7} - 1ight) \frac{1}{5} = \left(\frac{2}{7}ight) \frac{1}{5} = \frac{2}{35}$$

So, the new focal length is $f_{med} = 17.5	ext{ cm}$.

### The Final Calculation

With the new focal length, let's find the new image position. The object distance $u_2$ is still $-20	ext{ cm}$. Substituting into the lens formula:

$$\frac{1}{v_2'} - \frac{1}{-20} = \frac{2}{35}$$

$$\frac{1}{v_2'} = \frac{2}{35} - \frac{1}{20} = \frac{8 - 7}{140} = \frac{1}{140}$$

So, the new image is formed way out at $v_2' = 140	ext{ cm}$! The new magnification of the lens is $m_2' = \frac{140}{-20} = -7$.

The total magnification in the medium, $M_2$, is:

$$M_2 = m_1 	imes m_2' = (-2) 	imes (-7) = 14$$

Finally, the question asks for the magnitude of the ratio $\frac{M_2}{M_1}$. We simply divide $14$ by $2$, and we get our final answer:

$$\left|\frac{M_2}{M_1}ight| = \left|\frac{14}{2}ight| = \mathbf{7}$$

Notice how changing the surrounding medium drastically shifted the focal length of the lens, pushing the image further away and amplifying the magnification. A true masterpiece of optical physics!

Enter Numerical Value:

Visualized Solution

$Analyzing the Setup$

$Reflection at the Concave Mirror$

$Image Formed by the Mirror$

$Magnification of the Mirror$

$Refraction through the Lens (in Air)$

$Final Image in Air$

$Total Magnification (M_{1})$

$Immersing the Setup in a Medium$

$Calculating New Focal Length$

$Final Image in the Medium$

$Total Magnification (M_{2})$

$The Final Ratio$

The Sigma Insight: Lens

The Journey of Light

Reflection at the Concave Mirror

Refraction through the Convex Lens (in Air)

The Plot Twist

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Image Formed by the Mirror

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refraction through the Lens (in Air)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Magnification (M1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Immersing the Setup in a Medium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating New Focal Length

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

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

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