Question

A square digital image of size 128 × 128 pixels requires 65536 bits of storage. What is the number of grey levels?

• A. 16
• B. 4
• C. 8
• D. 32

Hint:

Find bits per pixel by dividing total storage bits by the total number of pixels, then use grey levels = 2^(bits per pixel).

Answer 1

The Correct Option is A

Step 1: Convert the total storage to bytes.
Total storage given is 65536 bits. Since 1 byte = 8 bits, \[ \text{Total bytes} = \frac{65536}{8} = 8192 \text{ bytes} \]

Step 2: Find bytes per pixel and hence bits per pixel.
The image has $128 \times 128 = 16384$ pixels. Bytes used per pixel: \[ \frac{8192}{16384} = 0.5 \text{ bytes/pixel} = 4 \text{ bits/pixel} \] This matches the direct bit calculation, confirming $k = 4$ bits per pixel.

Step 3: Use the grey level formula.
The maximum number of distinct grey levels representable with $k$ bits is $L = 2^k$ (each extra bit doubles the number of representable shades, from code 0 to code $L-1$). Substituting $k=4$, \[ L = 2^4 = 16 \]

Step 4: Sanity check against the other choices.
If $L$ were 4, 8 or 32 ($k=2,3,5$), the storage would work out to 32768, 49152 or 81920 bits respectively, none equal to the given 65536 bits. Only $L=16$ reproduces the stated storage exactly. \[ \boxed{L = 16} \]