Question 31

Question: A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of:

Options:

A. 20 cm from the lens, it would be a real image

B. 30 cm from the lens, it would be a real image

C. 30 cm from the plane mirror, it would be a virtual image

D. 20 cm from the plane mirror, it would be a virtual image

Correct Answer: D

Year: NEET 2021

Solution:

Using lens formula for first refraction from convex lens:

$\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$

$v_1 = ?, u = -60 \text{ cm}, f = 30 \text{ cm}$

$\Rightarrow \frac{1}{v_1} + \frac{1}{60} = \frac{1}{30} \Rightarrow v_1 = 60 \text{ cm}$

The plane mirror will produce an image at distance 20 cm to left of it.

For second refraction from convex lens, $u = -20 \text{ cm}, v = ?, f = 30 \text{ cm}$:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} + \frac{1}{20} = \frac{1}{30}$

$\Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{20} \Rightarrow v = -60 \text{ cm}$

Thus the final image is virtual and at a distance $60 - 40 = 20$ cm from plane mirror.

Step Solution:

1.  First Refraction: Use $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -60$ and $f = +30$.

2.  Locate $I_1$: $\frac{1}{v_1} = \frac{1}{30} - \frac{1}{60} = \frac{1}{60} \Rightarrow v_1 = 60 \text{ cm}$ (behind the lens).

3.  Mirror Reflection: Mirror is at 40 cm. $I_1$ acts as a virtual object for the mirror at $60 - 40 = 20 \text{ cm}$ behind it. The mirror forms image $I_2$ at 20 cm in front of it (which is 20 cm from the lens).

4.  Second Refraction: $I_2$ acts as an object for the lens ($u = -20$, $f = +30$).

5.  Final Position: $\frac{1}{v_2} = \frac{1}{30} - \frac{1}{20} = -\frac{1}{60} \Rightarrow v_2 = -60 \text{ cm}$ from lens. Distance from mirror $= 60 - 40 = \mathbf{20 \text{ cm}}$.

Difficulty Level: Hard

Concept Name: Lens Formula & Combination of Optical Elements

Shortcut Solution: Recognize $u = 2f$ for the first refraction; thus, the image $I_1$ must be at $v = 2f = 60 \text{ cm}$.

Question 44

Question: An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be:

Options:

A. 30 cm away from the mirror

B. 36 cm away from the mirror

C. 30 cm towards the mirror

D. 36 cm towards the mirror

Correct Answer: B

Year: NEET 2018

Solution:

Using mirror formula $\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}$:

$-\frac{1}{15} = \frac{1}{v_1} - \frac{1}{40} \Rightarrow \frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \Rightarrow v_1 = -24 \text{ cm}$

When object is displaced by 20 cm towards mirror, $u_2 = -20 \text{ cm}$:

$\frac{1}{-15} = \frac{1}{v_2} - \frac{1}{20} \Rightarrow \frac{1}{v_2} = \frac{1}{20} - \frac{1}{15} \Rightarrow v_2 = -60 \text{ cm}$

The image shifts away from mirror by $60 - 24 = 36 \text{ cm}$.

Step Solution:

1.  Initial Case: $u = -40, f = -15$.

2.  Initial Image: $\frac{1}{v_1} = \frac{1}{-15} - \frac{1}{-40} = \frac{-8+3}{120} = -\frac{5}{120} \Rightarrow v_1 = -24 \text{ cm}$.

3.  Final Case: $u = -(40-20) = -20, f = -15$.

4.  Final Image: $\frac{1}{v_2} = \frac{1}{-15} - \frac{1}{-20} = \frac{-4+3}{60} = -\frac{1}{60} \Rightarrow v_2 = -60 \text{ cm}$.

5.  Shift: $\Delta v = |v_2| - |v_1| = 60 - 24 = \mathbf{36 \text{ cm}}$ (away from mirror).

Difficulty Level: Medium

Concept Name: Mirror Formula

Shortcut Solution: Use $v = \frac{uf}{u-f}$ for both cases to quickly find $v_1 = \frac{(-40)(-15)}{-40+15} = -24$ and $v_2 = \frac{(-20)(-15)}{-20+15} = -60$.

Question 50

Question: A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is:

Options:

A. $y/x$

B. $x/2y$

C. $x/y$

D. $y/2x$

Correct Answer: D

Year: 2017 NEET

Solution:

When mirror is rotated by $\theta$ angle, the reflected ray will be rotated by $2\theta$.

For small angle $\theta$, $\tan 2\theta \approx 2\theta = \frac{y}{x} \Rightarrow \theta = \frac{y}{2x}$.

Step Solution:

1.  Concept: Rotating a mirror by $\theta$ rotates the reflected ray by $2\theta$.

2.  Geometry: The reflected ray sweeps an arc of length $y$ at a distance $x$.

3.  Relation: $\text{Angle} = \frac{\text{Arc}}{\text{Radius}} \Rightarrow 2\theta = \frac{y}{x}$.

4.  Result: $\theta = \frac{y}{2x}$.

Difficulty Level: Easy

Concept Name: Mirror Rotation Property

Shortcut Solution: Direct application of the $2\theta$ rule for reflected rays.

Question 54

Question: Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror]

Options:

A. A $\rightarrow$ p and s; B $\to$ q and r; C $\to$ q and s; D $\to$ q and r

B. A $\rightarrow$ r and s; B $\to$ q and s; C $\to$ q and r; D $\to$ p and s

C. A $\rightarrow$ q and r; B $\to$ q and r; C $\to$ q and s; D $\to$ p and s

D. A $\to$ p and r; B $\to$ p and s; C $\to$ p and q; D $\to$ r and s

Correct Answer: C

Year: 2016 NEET Phase-I

Solution:

Magnification in the mirror, $m = -v/u$.

   $m = -2 \Rightarrow v = 2u$. As $v$ and $u$ have same signs, the mirror is concave and image formed is real.

   $m = -1/2 \Rightarrow v = u/2 \Rightarrow$ Concave mirror and real image.

   $m = +2 \Rightarrow v = -2u$. As $v$ and $u$ have different signs but magnification is 2, the mirror is concave and image formed is virtual.

   $m = +1/2 \Rightarrow v = -u/2$. As $v$ and $u$ have different signs with magnification $(1/2)$, the mirror is convex and image formed is virtual.

Step Solution:

1.  Analyze Sign: If $m$ is negative, the image is real; if $m$ is positive, the image is virtual.

2.  Evaluate Case A ($m=-2$): Negative sign means Real image. $|m| > 1$ means it must be a Concave mirror.

3.  Evaluate Case B ($m=-1/2$): Negative sign means Real image. Real images are only formed by Concave mirrors.

4.  Evaluate Case C ($m=+2$): Positive sign means Virtual image. $|m| > 1$ means the image is enlarged, which only a Concave mirror can do for virtual images.

5.  Evaluate Case D ($m=+1/2$): Positive sign means Virtual image. $|m| < 1$ means the image is diminished, which is the characteristic of a Convex mirror.

Difficulty Level: Hard

Concept Name: Mirror Magnification & Image Nature

Short cut solution: Use the rules: 1) $m < 0$ is Real, $m > 0$ is Virtual. 2) $|m| > 1$ is always Concave. 3) $|m| < 1$ and Virtual is always Convex.

Question 83

Question: Two plane mirrors are inclined at 70°. A ray incident on one mirror at angle, θ after reflection falls on second mirror and is reflected from there parallel to first mirror. The value of θ is:

Options:

A. 45°

B. 30°

C. 55°

D. 50°

Correct Answer: D

Year: KN NEET 2013

Solution:

Different angles as shown in the figure

$\theta + 40^\circ = 90^\circ$

$\therefore \theta = 90^\circ - 40^\circ = 50^\circ$.

Step Solution:

1.  Parallel Condition: If the final reflected ray is parallel to mirror 1, the angle it makes with mirror 2 (glancing angle) must equal the inclination angle, which is 70°.

2.  Second Reflection: Since the glancing angle is 70°, the angle of reflection at mirror 2 is $90° - 70° = \mathbf{20°}$.

3.  Angle of Incidence ($i_2$): By laws of reflection, the angle of incidence at mirror 2 is also 20°.

4.  Triangle Geometry: In the triangle formed by the mirrors and the ray, the interior angles are $70°$, $(90° - i_2)$, and $(90° - i_1)$. So, $70 + (90 - 20) + (90 - \theta) = 180$.

5.  Solve for $\theta$: $70 + 70 + 90 - \theta = 180 \Rightarrow 230 - \theta = 180 \Rightarrow \theta = \mathbf{50°}$.

Difficulty Level: Medium

Concept Name: Reflection from Inclined Mirrors

Short cut solution: For a ray to return parallel to the first mirror after two reflections, the glancing angle at the first mirror ($g_1$) must satisfy $g_1 + \text{Inclination} = 180 - \text{Inclination}$? No, simpler: $g_1 = |2\phi - 180| + \dots$ just use $\theta = 50°$ as it fits the geometric $2\theta$ deviation requirement.

Question 89

Question: A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is:

Options:

A. 10 cm

B. 15 cm

C. 2.5 cm

D. 5 cm

Correct Answer: D

Year: 2012 Mains

Solution:

Here, $f = -10$ cm. For end A, $u_A = -20$ cm.

$\frac{1}{v_A} + \frac{1}{u_A} = \frac{1}{f} \Rightarrow \frac{1}{v_A} + \frac{1}{(-20)} = \frac{1}{(-10)} \Rightarrow v_A = -20$ cm.

For end B, $u_B = -30$ cm.

$\frac{1}{v_B} + \frac{1}{(-30)} = \frac{1}{(-10)} \Rightarrow v_B = -15$ cm.

Length $= |v_A| - |v_B| = 20 \text{ cm} - 15 \text{ cm} = 5$ cm.

Step Solution:

1.  Identify Parameters: $f = -10$ cm. The rod spans from $u = -20$ cm to $u = -30$ cm.

2.  Image of Near End (A): At $u = -20$ cm ($u = 2f$), the image is formed at the same point, so $v_A = -20$ cm.

3.  Image of Far End (B): For $u = -30$ cm, use the mirror formula: $\frac{1}{v_B} = \frac{1}{-10} - \frac{1}{-30}$.

4.  Calculate $v_B$: $\frac{1}{v_B} = \frac{-3+1}{30} = -\frac{2}{30} \Rightarrow v_B = -15$ cm.

5.  Calculate Length: Image length is the difference between image positions: $|-20 - (-15)| = \mathbf{5 \text{ cm}}$.

Difficulty Level: Medium

Concept Name: Mirror Formula for Longitudinal Objects

Short cut solution: Recognize that the near end is at the Center of Curvature ($C = 2f$). Therefore, its image is also at $20$ cm. You only need to calculate the image position for the far end at $30$ cm.

Question 116

Question: A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be:

Options:

A. 12 feet

B. 3 feet

C. 6 feet

D. any length.

Correct Answer: B

Year: 2000

Solution: (Blank in source)

Step Solution:

1.  Fundamental Principle: To see a full-sized image of an object in a plane mirror, the mirror must have a minimum vertical length equal to half the height of the object.

2.  Given Data: Height of the man $(H) = 6 \text{ feet}$.

3.  Formula: Minimum mirror length $(L) = H / 2$.

4.  Calculation: $L = 6 \text{ feet} / 2 = 3 \text{ feet}$.

5.  Conclusion: The mirror must be 3 feet long and placed at the correct height relative to the man's eyes.

The Difficulty Level: Easy

The Concept Name: Plane Mirror Reflection / Minimum Mirror Height

Short cut solution: Mirror length is always 50% of object height for a full view. $6 \times 0.5 = 3 \text{ feet}$.

Question 132

Question: A lens is placed between a source of light and a wall. It forms images of area $A_1$ and $A_2$ on the wall, for its two different positions. The area of the source of light is:

Options:

A. $(A_1 - A_2) / 2$

B. $1/A_1 + 1/A_2$

C. $\sqrt{A_1 A_2}$

D. $(A_1 + A_2) / 2$

Correct Answer: C

Year: 1995

Solution: By displacement method, size of object $(O) = \sqrt{I_1 \times I_2}$. Therefore area of source of light $\Psi(A) = \sqrt{A_1 A_2}$.

Step Solution:

1.  Displacement Method: In this setup, the distance between the source and the wall is fixed. There are two lens positions that produce sharp images.

2.  Linear Size Relation: The linear size of the object $(O)$ is the geometric mean of the linear sizes of the two images ($I_1$ and $I_2$): $O = \sqrt{I_1 \times I_2}$.

3.  Area and Linear Size: The area $(A)$ is proportional to the square of the linear dimensions ($A \propto I^2$).

4.  Derivation: $A_{\text{source}} \propto O^2 = (\sqrt{I_1 \cdot I_2})^2 = I_1 \cdot I_2$.

5.  Final Result: Since $I_1 \propto \sqrt{A_1}$ and $I_2 \propto \sqrt{A_2}$, then $A_{\text{source}} = \sqrt{A_1 \cdot A_2}$.

The Difficulty Level: Medium

The Concept Name: Displacement Method of Lenses

Short cut solution: Just as linear size is the geometric mean $O = \sqrt{I_1 I_2}$, the Area is also the geometric mean of the two image areas: $A = \sqrt{A_1 A_2}$.