Question

The largest wavelength of light falling on double slits separated by \(1.5\ \mu\text{m}\), for which there is a first-order maximum, is in the:

• A. Ultraviolet range
• B. Visible range
• C. Infrared range
• D. X-ray range

Hint:

Use \(d\sin\theta = \lambda\) with \(\sin\theta\) at most 1, so \(\lambda_{\max}=d\); then place it on the spectrum.

Answer 1

The Correct Option is C

Step 1: For a double slit the bright fringes satisfy \[d\sin\theta = m\lambda,\] where \(d\) is the slit separation and \(m\) is the order. Here \(m = 1\) (first-order maximum).
Step 2: So \(d\sin\theta = \lambda\). The wavelength is largest when \(\sin\theta\) is largest. Since \(\sin\theta \le 1\), the maximum possible value is \(\sin\theta = 1\).
Step 3: Therefore \[\lambda_{\max} = d = 1.5\ \mu\text{m} = 1500\ \text{nm}.\]
Step 4: Compare with the spectrum. Visible light is roughly \(400\text{-}700\ \text{nm}\); anything longer than \(700\ \text{nm}\) is infrared. Since \(1500\ \text{nm} > 700\ \text{nm}\), it lies in the infrared.
Step 5: The largest first-order wavelength is \(1500\ \text{nm}\), which is infrared. Answer (C). \[\boxed{\lambda_{\max} = 1.5\ \mu\text{m}\ \text{(infrared)}}\]