Two identical concave mirrors each of focal length f are facing each other as shown in the schematic diagram. The focal length f is much larger than the size of the mirrors. A glass slab of thickness t and refractive index n_0 is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source S is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on S itself, which of the following distances between the two mirrors is/are correct:

4f + (1 - \frac1n_0)t, 2f + (1 - \frac1n_0)t

Geometrical Optics PYQ — JEE Advanced 2025 · Shift 2 · Q38 with Solution

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Correct answer: A, B

Interpret the setup

Let the distance between the two identical concave mirrors be $D$ .

A glass slab of thickness $t$ and refractive index $n_0$ is placed symmetrically between them. The point source $S$ is at the center of the slab on the common principal axis.

Since $S$ is at the center of the slab, its distance from each slab surface is $t/2$ .

Also, because the slab is equidistant from the two mirrors, the physical air-gap from each mirror to the nearer slab face is the same. Let it be $a$ . Then $D = 2a + t.$

Effective optical distance from mirror to source

For paraxial imaging through a plane slab, a thickness $t$ of refractive index $n_0$ behaves like an equivalent air thickness $t/n_0$ for axial distance purposes.

Hence, the effective object distance of $S$ from either mirror is $u = a + \frac{t/2}{n_0}.$

Using $a = \frac{D-t}{2}$ ,

= \frac{D - t + t/n_0}{2} = \frac{D - \left(1-\frac1{n_0}\right)t}{2}.$$ Let $$\delta = \left(1-\frac1{n_0}\right)t.$$ Then $$u = \frac{D-\delta}{2}.$$ --- 3. **Condition for image to return to the source point** A ray from $S$ can return to $S$ after reflection from one mirror or after reflection from both mirrors. So we examine both possibilities. --- 4. **Case 1: Reflection from one mirror only** For a concave mirror, if the object is at the center of curvature, i.e. at distance $2f$, the image forms at the same point. Thus, for the image of $S$ due to either mirror alone to be formed back at $S$, $$u = 2f.$$ Substitute $u = \frac{D-\delta}{2}$: $$\frac{D-\delta}{2} = 2f$$ $$D - \delta = 4f$$ $$D = 4f + \delta$$ $$D = 4f + \left(1-\frac1{n_0}\right)t.$$ This is **Option A**. --- 5. **Case 2: Reflection from both mirrors successively** If after reflection from the left mirror, the rays become parallel, then after reaching the right mirror, the right mirror will focus those parallel rays at its focus. Because the setup is symmetric, that focus can coincide with $S$. For a concave mirror, parallel reflected rays occur when the object is at the focus. So we require $$u = f.$$ Then $$\frac{D-\delta}{2} = f$$ $$D - \delta = 2f$$ $$D = 2f + \delta$$ $$D = 2f + \left(1-\frac1{n_0}\right)t.$$ This is **Option B**. Indeed, in this case: - $S$ is at the focus of the left mirror, so after first reflection rays go parallel. - These parallel rays fall on the right mirror and are focused at its focus. - Since $S$ is also symmetrically at the focus of the right mirror, the final image is again at $S$. So Option B is also correct. --- 6. **Check Options C and D** Options C and D involve the slab correction as $(n_0-1)t$. But for axial imaging through a slab, the effective reduction in distance is $$t - \frac{t}{n_0} = \left(1-\frac1{n_0}\right)t,$$ not $(n_0-1)t$. Hence C and D are incorrect. --- 7. **Final result** The correct mirror separations are $$D = 4f + \left(1-\frac1{n_0}\right)t$$ and $$D = 2f + \left(1-\frac1{n_0}\right)t.$$

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View written solutionFree

Correct answer: A, B

Interpret the setup

Let the distance between the two identical concave mirrors be $D$ .

A glass slab of thickness $t$ and refractive index $n_0$ is placed symmetrically between them. The point source $S$ is at the center of the slab on the common principal axis.

Since $S$ is at the center of the slab, its distance from each slab surface is $t/2$ .

Also, because the slab is equidistant from the two mirrors, the physical air-gap from each mirror to the nearer slab face is the same. Let it be $a$ . Then $D = 2a + t.$

Effective optical distance from mirror to source

For paraxial imaging through a plane slab, a thickness $t$ of refractive index $n_0$ behaves like an equivalent air thickness $t/n_0$ for axial distance purposes.

Hence, the effective object distance of $S$ from either mirror is $u = a + \frac{t/2}{n_0}.$

Using $a = \frac{D-t}{2}$ ,

= \frac{D - t + t/n_0}{2} = \frac{D - \left(1-\frac1{n_0}\right)t}{2}.$$ Let $$\delta = \left(1-\frac1{n_0}\right)t.$$ Then $$u = \frac{D-\delta}{2}.$$ --- 3. **Condition for image to return to the source point** A ray from $S$ can return to $S$ after reflection from one mirror or after reflection from both mirrors. So we examine both possibilities. --- 4. **Case 1: Reflection from one mirror only** For a concave mirror, if the object is at the center of curvature, i.e. at distance $2f$, the image forms at the same point. Thus, for the image of $S$ due to either mirror alone to be formed back at $S$, $$u = 2f.$$ Substitute $u = \frac{D-\delta}{2}$: $$\frac{D-\delta}{2} = 2f$$ $$D - \delta = 4f$$ $$D = 4f + \delta$$ $$D = 4f + \left(1-\frac1{n_0}\right)t.$$ This is **Option A**. --- 5. **Case 2: Reflection from both mirrors successively** If after reflection from the left mirror, the rays become parallel, then after reaching the right mirror, the right mirror will focus those parallel rays at its focus. Because the setup is symmetric, that focus can coincide with $S$. For a concave mirror, parallel reflected rays occur when the object is at the focus. So we require $$u = f.$$ Then $$\frac{D-\delta}{2} = f$$ $$D - \delta = 2f$$ $$D = 2f + \delta$$ $$D = 2f + \left(1-\frac1{n_0}\right)t.$$ This is **Option B**. Indeed, in this case: - $S$ is at the focus of the left mirror, so after first reflection rays go parallel. - These parallel rays fall on the right mirror and are focused at its focus. - Since $S$ is also symmetrically at the focus of the right mirror, the final image is again at $S$. So Option B is also correct. --- 6. **Check Options C and D** Options C and D involve the slab correction as $(n_0-1)t$. But for axial imaging through a slab, the effective reduction in distance is $$t - \frac{t}{n_0} = \left(1-\frac1{n_0}\right)t,$$ not $(n_0-1)t$. Hence C and D are incorrect. --- 7. **Final result** The correct mirror separations are $$D = 4f + \left(1-\frac1{n_0}\right)t$$ and $$D = 2f + \left(1-\frac1{n_0}\right)t.$$

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