A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by

Q: Why can't a concave lens form a real image?

A concave lens is a diverging lens. For any real object placed in front of it, the refracted rays diverge and appear to meet on the same side as the object, always forming a virtual, erect, and diminished image.

Q: Why is the minimum distance between an object and its real image exactly $4f$?

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and applying sign convention ($u = -x$, $v = +y$), we get $\frac{1}{y} + \frac{1}{x} = \frac{1}{f}$. The total distance is $D = x + y$. Substituting $y = D - x$, we get a quadratic equation in $x$: $x^2 - Dx + Df = 0$. For real roots (real object position), the discriminant must be $\ge 0$, which means $D^2 - 4Df \ge 0$, leading to $D \ge 4f$.

Q: What if the distance $D$ was exactly equal to $4f$?

If $D = 4f$, the object must be placed exactly at $2f$, and the image will be formed exactly at $2f$ on the other side. In this specific case, the image is real and inverted, but it is of the **same size** as the object, not diminished.

Q: Why did we use strictly greater than ($>$) instead of greater than or equal to ($\ge$)?

The question specifically asks for a *diminished* image. If we used $\ge$, the equality case ($D = 4f$) would give an image of the same size. To ensure the image is diminished (or magnified), $u$ and $v$ must be unequal, which forces $D$ to be strictly greater than $4f$.

Q: Can a concave mirror also form a real diminished image?

Yes, a concave mirror can form a real diminished image if the object is placed beyond its center of curvature ($C$). However, a concave mirror was not among the given options.

Question

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by placing

Options:

(A) a plane mirror

(B) a convex mirror of suitable focal length

(C) a convex lens of focal length less than 0.25 m

(D) a concave lens of suitable focal length

Sigma Percentile · Accepted Answer

This problem is a beautiful test of your conceptual clarity regarding image formation by different optical instruments. Let's break it down step-by-step and see how logical deduction leads us straight to the answer. ### Analyzing the Setup The very first sentence of the question gives us a massive clue: the image is to be obtained on a **screen**. What does this imply? It means the image must be a **real image**. Virtual images, by definition, are formed by the apparent intersection of light rays and cannot be projected onto a screen. Now, let's evaluate the optical instruments given in the options: - **Plane Mirror:** Always forms a virtual, erect, and same-sized image for a real object. - **Convex Mirror:** Always forms a virtual, erect, and diminished image for a real object. - **Concave Lens:** Always forms a virtual, erect, and diminished image for a real object. By simple elimination, options (a), (b), and (d) are incorrect because they can never form a real image. This leaves us with only one candidate: the **convex lens**. ### The Master Equation A convex lens can form a real image. However, the question specifies that we need a **diminished** real image. For a convex lens, a real, inverted, and diminished image is formed when the object is placed beyond $2f$. In this case, the image is formed between $f$ and $2f$ on the other side of the lens. Let the object distance be $u$ and the image distance be $v$. The total distance between the object and the screen (where the image is formed) is $D = u + v$. From the fundamental properties of a convex lens, we know that the minimum distance between an object and its real image is exactly $4f$. This minimum distance occurs symmetrically when $u = 2f$ and $v = 2f$, resulting in an image of the same size as the object. ### Final Calculation Since we require a **diminished** image, the object and image distances cannot be equal ($u eq v$). Therefore, the total distance $D$ must be strictly greater than the minimum possible distance. Mathematically, we write this as: $$D > 4f$$ We are given that the distance between the object and the screen is $D = 1.0 ext{ m}$. Substituting this value into our inequality, we get: $$1.0 > 4f$$ Solving for the focal length $f$, we find: $$f < \frac{1.0}{4}$$ $$f < 0.25 ext{ m}$$ Thus, to achieve a diminished real image on a screen $1.0 ext{ m}$ away, we must use a **convex lens with a focal length strictly less than $0.25 ext{ m}$**. This perfectly matches option (c).

Animated Solution for Physics - Optics: A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by placing

Select Answer:

Visualized Solution

$Real Image Requirement$

$Eliminating Options$

$Convex Lens Properties$

$Minimum Distance Condition$

$Strict Inequality$

$Calculation$

$Final Conclusion$

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

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